Apply Rolle's theorem to find real roots 
Suppose the function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ such that $f(a)=f(b)=0$. Prove that there exist a point $c\in(a,b)$ such that
  $$f(c)-f'(c)=0$$
  From the question above, or otherwise, show that the equation
  $$1+x+\frac{x^2}{2!}+\cdots+\frac{x^{2n+1}}{(2n+1)!}=0$$
  has real roots on $\mathbb{R}$ but not more than two.
  Additionally, show that the equation
  $$e^x-x^n=0$$
  has at most three real roots on $\mathbb{R}$, where $n
\in \mathbb{N}$.

My attempt: Suppose the function
$$h(x)=e^{-x}f(x)$$
Then $$h'(x)=e^{-x}[f(x)-f'(x)]$$
Since $h(a)=h(b)=0$ and $e^{-x}>0$ for all real $x$, then there exist a point $c\in(a,b)$ such that $h'(c)=0$ i.e.  $f(c)-f'(c)=0$. Then I get stuck on the following question. Should I start it with construct a function
$$h(x)=e^{-x}[1+x+\frac{x^2}{2!}+\cdots+\frac{x^{2n+1}}{(2n+1)!}]$$
and follow my previous procedure?

Assume that there exist at least 4 distinct real roots of $e^x-x^n=0$. Then, by the first part of the question, we would have at least 3 distinct real points $c$ such that $e^c-nc^{n-1}=e^c-c^n$, or $nc^{n-1}=c^n$ (these points lie between neighboring roots of $e^x-x^n=0$). But the only roots of the equation $nc^{n-1}=c^n$ are $c=0$ and $c=n$, and we see that our assumption is wrong.  
