Can someone help me with this one? Prove by mathematical induction
For $$n\geq1$$ $$\displaystyle{\sum^n_ {k=0} k^n\binom{n}{k}(-1)^k= (-1)^nn!}$$
It's easy to see that for $$n=1$$ $$\displaystyle{0^1\binom{1}{0}(-1)^0+1^1\binom{1}{1}(-1)^1= -1}$$ and $$\displaystyle{(-1)^11!=-1}$$
My problem is how to use the induction hypothesis I'm trying to solve it this way:
Perhaps I just have to add this to my sum: $${\sum_{k={n+1}}^{n+1}} (n+1)^{n+1}\binom {n+1}{n+1}(-1)^{n+1} $$ And get: $$\displaystyle{\sum^n_ {k=0} \biggl[k^n\binom{n}{k}(-1)^k}\biggr]+(n+1)^{n+1}\binom {n+1}{n+1}(-1)^{n+1}$$ And as: $$\displaystyle{\sum^n_ {k=0} k^n\binom{n}{k}(-1)^k= (-1)^nn!}$$ Then i get: $$\displaystyle{(-1)^nn!+(n+1)^{n+1}(-1)^{n+1}}$$ I tried this: $$\left(-1\right)^{n}\,n!+\left(-n-1\right)\,\left(n+1\right)^{n}\, \left(-1\right)^{n}$$ And this: $$\left(-1\right)^{n}\,\left(n!-n\,\left(n+1\right)^{n}-\left(n+1 \right)^{n}\right)$$ But i can't figure out how to get to this: $$\displaystyle{ (-1)^{n+1}(n+1)!}$$
$\quad~$ Hint: Start with the famous $\displaystyle\sum_{k=0}^n{n\choose k}x^k=(1+x)^n$, then differentiate, and multiply with x. Repeat the process a few times, and see what happens. Deduce and prove by induction the formula that should be pretty evident by now, and then let the number of repeated differentiations followed by multiplications be n, and x be $-1$. QED.