Numerical differentiation with Binomial Theorem In George Shilov's Elementary Real and Complex Analysis, there is a problem which asks us prove

If $f$ is twice differentiable on some open interval and the second derivative is continuous at $x$, then prove that
  $$f''(x)=\lim_{h\rightarrow 0}\frac{f(x)-2f(x+h)+f(x+2h)}{h^2}\,.$$

This is a common fact in numerical differentiation to approximate derivatives at the left-hand point and is fairly immediate from two applications of Taylor's Theorem with Lagrange Remainder. However, this was not the end of Shilov's problem. He also states

Find a similar expression for $f^{(n)}(x)$ (with appropriate hypotheses).

In the back of his book, he asserts that

$$f^{(n)}(x)=\lim_{h\rightarrow 0}\frac{1}{h^n}\sum_{k=0}^n (-1)^k\binom{n}{k}f(x+kh)$$

which I found interesting enough to at least remember, if not attempt. However, I recently came upon an application where this formula would be useful and attempted to prove it. However, it seems there was an error in Shilov's claim. He must have meant

$$(-1)^nf^{(n)}(x)=\lim_{h\rightarrow 0}\frac{1}{h^n}\sum_{k=0}^n (-1)^k\binom{n}{k}f(x+kh)$$

because working out $n=3$ and applying Lagrange's Remainder three times results in

$$\frac{f(x)-3f(x+h)+3f(x+2h)-f(x+3h)}{h^3}=\frac{1}{3!}\left(-3f'''(\xi_1)+24f'''(\xi_2)-27f'''(x_3)\right)$$

which gives the corrected limit (with continuity of $f^{(3)}$ at $x$ assumed).

Is there an easy way to go about proving this result in general?

We can attack this fairly directly, without induction. But this becomes equivalent to proving several interesting binomial identities:

$$\sum_{k=0}^n(-1)^k\binom{n}{k} k^m=\begin{cases} (-1)^n n!&\text{ if }m=n\\0&\text{ if }0\leq m<n\end{cases}$$

The first of which was tackled here while the others seem to have gone largely unasked. The case $m=1$ is tackled here and here, and I can see that I could continue the approaches taken in these answers by differentiating several times. The book-keeping isn't too awful because all these identities are just sums of $0$s. Thus $0\leq m<n$ isn't too bad, if we can do $m=1$. However, proving the cases $m=1$ and $m=n$ aren't entirely trivial.
Shilov seems to have hidden an interesting exercise in a terse sentence without any hint that it would be interesting. This makes me wonder if there's an easier way to go about proving this result.

I'm not sure I'm contributing anything, maybe I misunderstood since this feels like essentially a duplicate of this link.  $\newcommand{\fd}{\Delta}$ Its not so hard (via e.g. repeated applications of l'Hopital, as Paramanand shows in that link) to show that you are actually interested in iterated forward finite differences
$$ \fd_h^1 f(x) := f(x+h)-f(x), \quad \fd_h^{n+1} f(x) := \fd_h^1[\fd_h ^nf] (x)=\fd_h ^nf(x+h) - \fd_h ^nf(x)$$
And your hope is that 
$$ \fd_h ^nf (x) = \sum_{k=0}^{n} (-1)^{n-k} \binom{n}k f(x+kh)$$
By rescaling $f$, and choosing a different $x$, it would suffice to do this for $h=1$; let $\fd:=\fd^1_1$ for easy typing. Now the inductive proof for this is  straightforward using the Pascal triangle,
\begin{align} \fd^{n+1}f(x)&=\Delta\left[\sum_{k=0}^{n} (-1)^{n-k} \binom{n}k f(\bullet +k)\right](x) \\ &= \sum_{k=0}^{n} (-1)^{n-k} \binom{n}k f(x +k+1) - \sum_{k=0}^{n} (-1)^{n-k} \binom{n}k f(x+k)\\
&= \sum_{k=1}^{n+1} (-1)^{n+1-k} \binom{n}{k-1} f(x +k) + \sum_{k=0}^{n} (-1)^{n+1-k} \binom{n}k f(x+k)\\
&=(-1)^{n+1}  f(x) + f(x+n+1) + \sum_{k=1}^n (-1)^{n+1-k} \underbrace{\left(\binom{n}{k-1} +\binom{n}{k}\right)}_{=\binom{n+1}{k}}f(x +k)\\
&=\sum_{k=0}^{n+1} (-1)^{n+1-k} \binom{n+1}{k}f(x +k) \end{align}
