I'm really stuck on this question
Let $f(x)$ be a function on $[a,b]$ which is $(n + 1)$ times differentiable at every point, for some natural number $n$. Let $c$ be a point in $(a,b)$. Using induction and the generalized mean-value theorem prove that for any $x \in [a,b] \setminus {c}$,
$$f(x)-\sum_{n,k=0}\frac{f^k(c)}{k!(x-c)^k}\frac{(x-c)^{n+1}}{(n+1)!} = f^{(m)}(x_m)-\sum_{n-m,k=0}\frac{f^{(k+m)}(c)}{k!(x_m-c)^k}\frac{(x_m-c)^{(n+1-m)}}{(n+1-m)!}$$
for some $x_m$ which lies between $x$ and $c$, for $m = 1, \dots, n + 1$. By applying this with $m = n + 1$ deduce that
$$f(x)=f(c)+f'(c)(x-c)+... [\frac{f^{(n)}(c)}{n!}](x-c)^n+[\frac{f^{(n+1)}(\theta)}{(n+1)!}](x-c)^{(n+1)}$$
I've tried to prove by induction by proving true for n=1 and m=1. Ended up with equation below
$$\frac{f(x)-f'(c)(x-c)-f(c)}{\frac{(x-c)^2}{2}}=\frac{f'(x_1 )-f'(c)}{x_1-c} $$