So the first bit of the question gives you the approximation...

$\displaystyle f(a+h)\approx f(a)+f^{(1)}(a)h+\frac{f^{(2)}(a)h^2}{2!}+\frac{f^ {(3)}(a)h^3}{3!}+\frac{f^{(4)}(a)h^4}{4!}$ with $\displaystyle f(x)=e^x$ and $\displaystyle a=0$

...to find the approximate value of $\displaystyle e^1$, which I've done.

The the second part is...

The Extended Mean Value Theorem says that for fixed $\displaystyle f$, $\displaystyle a$ and $\displaystyle h$, there is a $\displaystyle \theta$ in the range $\displaystyle 0\leq\theta\leq1$, such that

$\displaystyle f(a+h)=$$\displaystyle f(a)+f^{(1)}(a)h+\frac{f^{(2)}(a)h^2}{2!}+\frac{f^ {(3)}(a)h^3}{3!}+\frac{f^{(4)}(a)h^4}{4!}+\frac{f^ {(5)}(a+\theta h)h^5}{5!}$

Use this to show that for some $\displaystyle \theta '$ with $\displaystyle 0\leq\theta '\leq1$ we have

$\displaystyle 4!e^1=65+\frac{3\theta '}{5}$

...which I think I've managed okay, but it's this next bit that I'm stuck on...

Deduce more generally that

$\displaystyle \color{red}n!e^1=N_n+\frac{3\theta_n}{(n+1)}$

...for some integers $\displaystyle \color{red}N_n$ and numbers $\displaystyle \color{red}\theta_n$ where $\displaystyle \color{red}0\leq\theta_n\leq1$.

Can anyone help? I just can't get my head around it.