Here is another proof of Lagrange's estimate using repeated application of Rolle's theorem.
Let $\displaystyle x\not =0$ and let $\displaystyle a$ be the solution to the equation:
$\displaystyle f(x)=\sum_{k=0}^{n}\frac{f^k(0)}{k!}x^k + \frac{ax^{n+1}}{(n+1)!}$.
The argument will be complete if we can show $\displaystyle a=f^{(n+1)}(\mu)$ for some $\displaystyle \mu$ between $\displaystyle 0$ and $\displaystyle x$.
Now define the function,
$\displaystyle g(y) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!}y^k + \frac{ay^{n+1}}{(n+1)!} - f(y)$.
Notice that $\displaystyle g(0)=g'(0)=g''(0)=...=g^{(n)}(0)=0$.
Since $\displaystyle g(x) = 0$ by definition of $\displaystyle a$ we know by Rolle's theorem there is $\displaystyle x_1$ between $\displaystyle 0$ and $\displaystyle a$ so that $\displaystyle g'(x_1)=0$. But since $\displaystyle g'(0)=g'(x_1)$ by Rolle's theorem again there is an $\displaystyle x_2$ between $\displaystyle 0$ and $\displaystyle a$ so that $\displaystyle g''(x_2)=0$. This continues until we get an $\displaystyle x_{n+1}$ between $\displaystyle 0$ and $\displaystyle x_n$ so that $\displaystyle g^{(n+1)}(x_{n+1})=0$. But then by definition of $\displaystyle g$ it follows that $\displaystyle a = f^{(n+1)}(x_{n+1})$, choose $\displaystyle \mu = x_{n+1}$ and the proof is complete.