Proving formula for $n$-th derivative using l'Hopital (given existence of $n$-th derivative) 
Let $f:J\rightarrow\mathbb{R}$ be $n$-times differentiable on an open
interval $J$  ($f^{(n)}$ is not necessarily continuous) and $x\in J$.
Then  $$\lim_{h\rightarrow
 0}h^{-n}\sum_{i=0}^n\binom{n}{i}(-1)^{n-i}f(x+ih)=f^{(n)}(x).$$

I think I have to use induction + l'hopital. Let $n=2$. Then
$$h^{-2}\sum_{i=0}^2\binom{2}{i}(-1)^{2-i}f(x+ih)=h^{-2}(f(x)-2f(x+h)+f(x+2h).$$
Differentiating the above w.r.t $h$ once yields
$$\frac{2f'(x+2h)-2f'(x+h)}{2h}=\frac{f'(x+2h)-f'(x+h)}{h}.$$
Define $g_1(h):=\frac{f'(x+h)-f'(x)}{h}-f''(x)$ and $g_2(h):=\frac{f'(x+2h)-f'(x)}{2h}-f''(x)$. Clearly $g_1(h),g_2(h)\rightarrow 0$ as $h\rightarrow 0$ and
$$\frac{f'(x+2h)-f'(x+h)}{h}=f''(x)+2g_2(h)-g_1(h).$$
Then taking $h\rightarrow 0$ we get the desired result.
Now suppose the result holds for $n$. How can I generalize the above method to proceed?
Edit: It seems that I need to prove that for all $n$ there exist functions $g_1(h),\ldots,g_n(h)$ such that $\lim_{h\rightarrow 0}g_i(h)=0$ and
$$\frac{\frac{d^{n-1}}{dh^{n-1}}\sum_{i=0}^n\binom{n}{i}(-1)^{n-i}f(x+ih)}{n!h}=f^{(n)}(x)+\sum_{i=1}^ng_i(h).$$
Now, supposing the above holds for $n$. How can I prove that it holds for $n+1$?
 A: Induction is bad since it hides what is going on. Recognize $\sum_{i = 0}^{n}\binom{n}{i}(-1)^{n - i}f(x + ih)$ as a result of applying the binomial theorem. To see this, write
$$\sum_{i = 0}^{n}\binom{n}{i}(-1)^{n - i}f(x + ih) = \sum_{i = 0}^{n}\binom{n}{i}(-1)^{n - i}(\tau_h^if)(x), \hspace{20pt} (\tau_hg)(x) = g(x + h).$$
So we should investigate the operator
$$\sum_{i = 0}^{n}\binom{n}{i}(-1)^{n - i}\tau_h^i = \sum_{i = 0}^{n}\binom{n}{i}\tau_h^i(-I)^{n - i}.$$
The binomial theorem gives
$$\sum_{i = 0}^{n}\binom{n}{i}\tau_h^i(-I)^{n - i} = (\tau_h - I)^n.$$
Therefore you want to compute
$$\lim_{h \to 0}\left(\frac{(\tau_h - I)^n}{h^n}f\right)(x) = \lim_{h \to 0}\left(\left(\frac{\tau_h - I}{h}\right)^nf\right)(x) = \lim_{h \to 0}(\Delta_h^nf)(x),\hspace{20pt}(\Delta_hg)(x) = \frac{g(x + h) - g(x)}{h}.$$
It remains to show this limit is $f^{(n)}(x)$. By the mean value theorem ($(\Delta_hg)(x) = g'(x + \theta(h)h)$, $\theta(h) \in (0, 1)$), we get
$$(\Delta_h^nf)(x) = (D\Delta_h^{n - 1}f)(x + \theta(h)h) = (\Delta_h^{n - 1}Df)(x + \theta(h)h) = (\Delta_h^{n - 1}f')(x + \theta(h)h).$$
Iterating this, we get
$$(\Delta_h^{n - 1}f')(x + \theta(h)h) = (\Delta_hf^{(n - 1)})(x + \theta_1(h)h + \dots + \theta_{n - 1}(h)h).$$
Now you can use the definition of derivative as a best linear approximation ($g(x + y) = g(x) + g'(x)y + yR(y)$, $R(y) \to 0$ as $y \to 0$) to show that this converges to $f^{(n)}(x)$.
