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?
 A: Your polynomial $p(x)$ has odd degree so it must have at least one root $a$.
You can observe that for every $x\geq 0$ the polynomial $p(x)>0$, so the roots of the polynomial are all negative.
If there is another root $b$ (you can suppose WLOG that $a\leq b$) you have that $p(a)=p(b)=0$ so there exist $c\in (a,b)\subset (-\infty, 0)$ such that 
$ p(c)-p’(c)=\frac{c^{2n+1}}{(2n+1)!}=0$
And it is not possible .
A: The answer above addressed the second part of your question, so I guess I'll answer the third.  We break it into cases for $n$ even and $n$ odd.  
Let $f(x)=e^x-x^n$, so that $f'(x)=e^x-nx^{n-1}$
First, assume $n$ is odd, so that $x<0\rightarrow x^n<0$; this along with the fact that $0$ is not a root of $f$ means our roots must be positive.  Assume we have two roots $a \neq b$ and assume w.l.o.g that $a<b$, from the above result, we have that $\exists_{c\in\mathbb{R}}f(c)=f'(c)$ so we have that $e^c-c^n=e^c-nc^{n-1}\rightarrow c=n$,and obviously $c$ can only belong to one interval of roots of $f$.
Now, let $n$ be even, we apply the exact same reasoning as above to show that $f$ can have at most $2$ positive roots.  To show that there can be at most $1$ negative root, we note that $x^n$ is strictly decreasing for $x\in\mathbb{R}^{-}$ and $e^x$ is strictly increasing for $x\in\mathbb{R}$ so they can (and do by IVT) intersect at most once.
So then, the sum of roots is at most $3$ for even $n$ and at most $2$ for odd $n$, so is at most $3$.
A: 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.  
