A general expression for the $n$th derivative Can someone help me prove the following
$$f^{(n)}(x_0)=\lim_{h \rightarrow 0}\frac{1}{h^{n}}\sum\limits_{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)(-1)^{n-k}f(x_0+kh)
$$
I have managed to manipulate the above expression with l'Hôpital's rule, but to no avail.
Thanks.
 A: Here is an amusing proof.
For $k=0,\dots ,n$, consider the "Taylor expansion of order $n$" for $f(x_0+kh)$ :
$$f(x_0+kh)=\sum_{i=0}^n \frac{f^{(i)}(x_0)}{i!}k^ih^i+o(h^n) $$
Then write down $\sum\limits_{k=0}^n \left(\begin{matrix}n\\k\end{matrix}\right)(-1)^{n-k} f(x_0+kh)$ and interchange the summations : this gives 
\begin{equation}\sum_{k=0}^n \left(\begin{matrix}n\\k\end{matrix}\right)(-1)^{n-k} f(x_0+kh)=\sum_{i=0}^n c_i f^{(i)}(x_0)h^i +o(h^n)\,, \tag{$*$}\end{equation}
where the coefficients $c_i$ are $independent\;of\; f\;and\; h$ (this is the key point).
Next, consider the function $F(x)=e^x$. By a direct computation (using the binomial formula) we get
\begin{eqnarray*}
\sum_{k=0}^n \left(\begin{matrix}n\\k\end{matrix}\right)(-1)^{n-k} F(x_0+kh)&=&e^{x_0}\sum_{k=0}^n \left(\begin{matrix}n\\k\end{matrix}\right)(-1)^{n-k}e^{kh}\\
&=&e^{x_0}\left(e^{h}-1\right)^n\\
&=&e^{x_0}h^n+o(h^n)\, .
\end{eqnarray*}
But, by formula ($*$) above, we should also have
\begin{eqnarray*}
\sum_{k=0}^n \left(\begin{matrix}n\\k\end{matrix}\right)(-1)^{n-k} F(x_0+kh)&=&
\sum_{i=0}^n c_iF^{(i)}(x_0)h^i+o(h^n)\\
&=&\sum_{i=0}^n c_ie^{x_0} h^i +o(h^n)
\end{eqnarray*}
Identifying the coefficients, it follows that $c_i=0$ for $i=0,\dots ,n-1$ and $c_n=1$. Going back to formula ($*$), we conclude that
$$\sum_{k=0}^n \left(\begin{matrix}n\\k\end{matrix}\right)(-1)^{n-k} f(x_0+kh)=f^{(n)}(x_0) h^n +o(h^n) $$ 
for $any$ (smooth) function $f$. Dividing by $h$, this gives the required limit.
