Prove that $\sum_{x=0}^{n}(-1)^x\binom{n}{n-x} (n+1-x)^n=n!$ I figure out these thing when "playing" with numbers:
$$3^2-2.2^2+1^2=2=2!$$
$$4^3-3.3^3+3.2^3-1^3=6=3!$$
$$5^4-4.4^4+6.3^4-4.2^4+1^4=24=4!$$
So I go to the conjecture that:
$$\binom{n}{n}(n+1)^n-\binom{n}{n-1}n^n+\binom{n}{n-2}(n-1)^n-...=n!$$
or
$$\sum_{x=0}^{n}(-1)^x\binom{n}{n-x} (n+1-x)^n=n!$$
Now, how can I prove this conjecture? I've tried a lot, but still couldn't have any idea.

First of all note that $$\sum_{k=0}^{n}\dbinom{n}{n-k}\left(-1\right)^{k}\left(n-k+1\right)^{n}=\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\left(n-k+1\right)^{n}$$ then from the special case of the Melzak's identity: $$\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\frac{f\left(y-k\right)}{x+k},=\frac{f\left(x+y\right)}{x\dbinom{x+n}{n}}-n!a_{n+1}\,x,y\in\mathbb{R},\,x\neq-k$$ where $f $ is an algebraic polynomial of degree $n+1$ and $a_{n+1}$ is the coefficient of the $n+1-$th power, we have, taking $f\left(z\right)=\left(z+1\right)^{n+1},\,y=n$ $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{\left(n-k+1\right)^{n+1}}{-x-k}=n!-\frac{\left(n+x+1\right)^{n+1}}{x\dbinom{x+n}{n}}$$ and so $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\left(n-k+1\right)^{n}=\lim_{x\rightarrow-n-1}\left(n!-\frac{\left(n+x+1\right)^{n}}{x\dbinom{x+n}{n}}\right)={\color{red}{n!}}$$ as wanted.
