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

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
$$ \frac{d^n}{dx^n} \Bigg( \sum_{j=0}^{n} \frac{x^j}{j!} \Bigg)^k \Bigg|_{x=0}
= \frac{d^n}{dx^n} (e^x)^k |_{x=0}
= k^n e^{kx} |_{x=0}
= k^n.$$
