Hi I do not understand the proof

Theorem. (Hermite 1873). The number e, the base of natural logarithms,is transcendental.

Proof. (1) Use integration by parts on the integral $\displaystyle \displaystyle\int_{0}^{x} e^{-t}f(t)\, dt$ andmultiply the resulting equality by $\displaystyle e^x $ to obtain:

$\displaystyle e^x \displaystyle\int_{0}^{x} e^{-t}f(t)\, dt = e^xf(0)-f(x)+ e^x\displaystyle\int_{0}^{x} e^{-t}f^{\prime}(t)\, dt $

Note that the integral on the right-hand side is obtained from the one onthe left-hand side by replacing f by f'. This means that if we write thesame identity for successive derivatives f' , f'', f''', etc. in place of f and addthe results, there will be cancellations. In particular, assume now f is apolynomial. Define:

$\displaystyle F(x) = \displaystyle\sum^{\infty}_{\substack{j=0 }}F^{(j)}(x)$ ,

which is a finite sum for polynomial f. Then (after cancellation) the resultof adding the identities above for the functions f, f' , f'', etc. can be writtenusing F in the form:

$\displaystyle e^x \displaystyle\int_{0}^{x} e^{-t}f(t)\, dt = e^xF(0)-F(x) $(*)

What are we using about e here? The fact that the function $\displaystyle e^x $ has the value1 at x = 0 and is equal to its own derivative. (It is the only differentiablefunction with this property.)

(2) Now assume (by contradiction) e is algebraic. By definition, thismeans there is a polynomial p(t) with integer coefficients $\displaystyle a_j(a_0 \neq{0} )$ anddegree $\displaystyle n\geq 1$ , so that $\displaystyle p(e) = 0 $:

$\displaystyle a_0+a_1e+a_2e^2+...+a_ne^n=0$ (1)

(Note we allow n = 1, so in particular we are not assuming a priori that e isirrational, a fact which had been proved by Liouville about 20 years earlier.)Write the identity (*) for x = k = 0, 1, 2, . . . , n, multiply by ak and addthe results. We obtain

$\displaystyle \displaystyle\sum_{k=0}^n{a_ke^k} \displaystyle\int_{0}^{k} e^{-t}f(t)\, dt = F(0) \displaystyle\sum_{k=0}^n{a_ke^k}- \displaystyle\sum_{k=0}^n{a_kF(k)} $

Using the equation (1), we may re-write this in the form:

$\displaystyle \displaystyle\sum_{k=0}^n{a_k F(k)}= - \displaystyle\sum_{k=0}^n{a_ke^k} \displaystyle\int_{0}^{k} e^{-t}f(t)\, dt$ (**)

(3) Note that, so far, we are still free to choose the polynomial f(t). Theidea of the proof now is to choose f so that the left-hand side of (**) isa non-zero integer (hence greater than or equal to one in absolute value),while the right-hand side is small, giving a contradiction. Hermite’s inspiredchoice for f is:

$\displaystyle f(t) = \displaystyle\frac{1}{(p-1)!} t^{p-1}g(t)^p$ donde $\displaystyle g(t)=(t-1)(t-2)...(t-n)$(2)

and p is a prime number that we’re free to choose as large as needed. Denoteby A the maximum absolute value of tg(t), over the interval [0, n]. Then wehave for the right-hand side of (**) the bound: (I DO NOT UNDERSTAND THE BOUND)

$\displaystyle | \displaystyle\sum_{k=0}^n{a_ke^k} \displaystyle\int_{0}^{k} e^{-t}f(t)\, dt | \leq{\displaystyle\frac{1}{(p-1)!}} (\displaystyle\sum_{k=0}^n{| a_k| })e^nnA^p $ (I DO NOT UNDERSTAND THIS PART)

where $\displaystyle A p / ( p - 1)! $ is an upper bound for the integrand over the interval ofintegration [0, n]). On the right-hand side of this estimate, the coefficientsak, the degree n and the number A are fixed, independent of p. Since thefactorial function grows faster than any exponential, this implies the righthandside of (**) can be made as small as desired (say, less than 1/2), bychoosing the prime p large enough.

(4) To conclude the proof, we just need to check that $\displaystyle \displaystyle\sum_{k=0}^n{a_kF(k)} $ isa nonzero integer. Here F(k) is the sum of the values of f and all itsderivatives at an integer k, and the polynomial g has integer coefficients, sothe denominator $\displaystyle ( p - 1)!$ in f is our potential problem. We first check whathappens at k = 0. Since t = 0 is a zero of f(t) of multiplicity p − 1, theTaylor expansion of f at t = 0 (which is really a finite sum, since f is apolynomial) has the form:

$\displaystyle f(t)= \displaystyle\frac{1}{(p-1)!}f^{(p-1)}(0)t^{p-1}+\displaystyle\frac{1}{p!}t^pf^p(0)t^p+...+ \displaystyle\frac{1}{j!}f^{(j)}(0)t^j+..$ (3)

Clearly $\displaystyle f^{(j)}(0)=0 $ for $\displaystyle j < p - 1$ and comparing coefficients of $\displaystyle t^{ p- 1} $ in (2) and (3) we see that $\displaystyle f^{ ( p- 1)} (0) $ is the $\displaystyle p^{th}$ power of the constant term in thepolynomial g(t):

$\displaystyle f ^{( p- 1)} (0) = {[( - 1)^ n n !]}^ p $

So $\displaystyle f^{( p- 1)} (0)$ is an integer. At this point we impose the last largeness requirementson p :$\displaystyle p > n$ and p $\displaystyle > | a_0 |$ (The reason for the second requirement willbe seen below.) Since p is a prime number greater than n, it does not occurin the prime factorization of n!. Hence we know that the integer $\displaystyle f^{( p- 1)} (0) $ is not a multiple of p. As for the higher-order derivatives of f at zero, againcomparing coefficients of powers of t in (2) and (3) we see that, for j $\displaystyle \geq $ p

$\displaystyle f^{(j)}(0)= \displaystyle\frac{j!}{(p-1)!} \times{coeficient}$ ( of $\displaystyle t^{j-(p-1)}$ in $\displaystyle g(t)^p $),

and the quantity in parenthesis is certainly an integer. Since $\displaystyle j\geq p $ , the number $\displaystyle j ! / ( p - 1)! $ is also an integer, and a multiple of p. This shows F(0)is a non-zero integer (since it is the sum of integers, only one of which is nota multiple of p), and likewise for $\displaystyle a_0F(0) $

(5)The proof that F(k) is an integer for k = 1, 2, . . . n is similar, butconsidering Taylor expansions at k. Take, for example k = 2 (if n $\displaystyle \geq 2$) Since 2 is a zero of f with multiplicity p, certainly $\displaystyle f^(j) (2) = 0 $ for $\displaystyle j < p$ , y we have:

$\displaystyle f(t)=\displaystyle\frac{1}{p!}f^{(p)}(2){(t-2)}^p+...+\displaystyle\frac{1}{j!}f^{(j)}(2){(t-2)}^j+...$ (4)

Comparing coefficients in (2) and (4), we find, for j $\displaystyle \geq$ p

$\displaystyle f^{(j)}(2)=\displaystyle\frac{j!}{(p-1)!}$ (coefficient of $\displaystyle t^{j-(p-1)}$ in $\displaystyle g(t)^p $ ),

(By a change of variable, we can always express the polynomial $\displaystyle t^{(p-1)}g(t)^p$ in terms of powers of (t − 2), instead of powers of t.) Note that, for j $\displaystyle \geq $ p this clearly shows $\displaystyle f^{(j)}(2) $ ) is an integer, and a multiple of p. So F(2) is alsoan integer, and multiple ofPp; and likewise for k = 1, 3, . . . n, showing that $\displaystyle \displaystyle\sum_{k=1}^n{a_kF(k)} $ is an integer, and multiple of p; and recalling the result ofpart (4) we even have that the sum starting at zero, $\displaystyle S= \displaystyle\sum_{k=0}^n{a_kF(k)} $ , is an integer.

Thanks