How to prove that $k^n=\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})^k|_{x=0}$ $$k^n=\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})^k|_{x=0}$$
It is easy to show $k=1$ and $k=2$
$k=1$
$$\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})|_{x=0}=1$$
$k=2$
$$\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})|_{x=0}$$
If we collect $x^n$ terms;
$$\frac{d^n}{dx^n} (1+2x+....+(\frac{n!}{0!n!}+\frac{n!}{1!(n-1)!}+...+\frac{n!}{n!0!})\frac{x^n}{n!}+.....)|_{x=0}$$
$$\frac{d^n}{dx^n} (1+2x+....+(\frac{n!}{0!n!}+\frac{n!}{1!(n-1)!}+...+\frac{n!}{n!0!})\frac{x^n}{n!}+.....)|_{x=0}=(\frac{n!}{0!n!}+\frac{n!}{1!(n-1)!}+...+\frac{n!}{n!0!})=2^n$$
I thought to use Leibniz's  method to prove general case but it seems we need to know $$\frac{d^{n-r}}{dx^{n-r}} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})^k|_{x=0}$$
Therefore I am stuck . 
Is there an easy way to prove the general case?
Thanks a lot for helps
 A: Notice that
$$ \frac{d^n}{dx^n} \Bigg( \sum_{j=0}^{n} \frac{x^j}{j!} \Bigg)^k \Bigg|_{x=0} = 
\frac{d^n}{dx^n} \Bigg( \sum_{j=0}^{n} \frac{x^j}{j!} + x^{n+1}f(x) \Bigg)^k \Bigg|_{x=0}$$
for any $n$-times differentiable function $f$. So we may choose $f$ at our convenience. Now if we choose
$$f(x) = \sum_{j=n+1}^{\infty} \frac{x^{j-n-1}}{j!}$$
then we get
$$ \frac{d^n}{dx^n} \Bigg( \sum_{j=0}^{n} \frac{x^j}{j!} \Bigg)^k \Bigg|_{x=0}
= \frac{d^n}{dx^n} (e^x)^k |_{x=0}
= k^n e^{kx} |_{x=0}
= k^n.$$
A: Let 
$$ s_n = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+…+\frac{x^n}{n!} $$
We can use induction on $n$. $n=1$ is easy to prove. Assume the hypothesis is true for $1 .. (n-1)$.
We have
$$ \frac{d}{dx}s_n = s_{n-1}$$
Now:
$$ \frac{d^n}{dx^n}s_n^k = \frac{d^{n-1}}{dx^{n-1}}(\frac{d}{dx}s_n^k) $$
And, using the chain rule:
$$ \frac{d}{dx}s_n^k = k\cdot s_n^{k-1} \cdot s_{n-1}$$ 
$$ = k\cdot (s_{n-1}+\frac{x^n}{n!})^{k-1} \cdot s_{n-1}$$
Using the Binomial Theorem:
$$ = k\cdot \sum_{i=0}^{k-1}{{k-1} \choose i} \frac{x^{ni}}{n!^{i}} s_{n-1}^{k-1-i}\cdot s_{n-1}$$
$$ = k\cdot \sum_{i=0}^{k-1}{{k-1} \choose i}\frac{x^{ni}}{n!^{i}} s_{n-1}^{k-i}$$
Taking the $n-1^{th}$ derivative of these terms, we can see that all the terms for $i>0$ will be zero at $x=0$ because of the factor $x^{ni}$. The only term left will be $i=0$. Therefore:
$$ \frac{d^n}{dx^n}s_n^k = k \cdot \frac{d^{n-1}}{dx^{n-1}} s_{n-1}^k = k^n$$
