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.

We want to answer to this question with combinatoric method. Suppose that we have $n$ elements $(x_1,x_2,\cdots,x_n)$. We know that all permutation of these elements are $n!$ and now we want to count this $n!$ cases with the following method. 
Consider the $n+1$ elements $(x_1,x_2,\cdots,x_n,y)$ and let we want to fill $n$ places with these $n+1$ elements. All possible cases that these $n$ places can be filled with this $n+1$ elements are $(n+1)^n$ but we should remove some cases from these $(n+1)^n$ cases to just remain the permutation of $(x_1,x_2,\cdots,x_n)$.
The first case is that we choose $n-1$ element of $(x_1,x_2,\cdots,x_n)$ like $(x_1,x_2,\cdots,x_{n-1})$ and after that remove all possible cases that the $n$ places can be filled with the $(x_1,x_2,\cdots,x_{n-1},y)$ that numbers of the first case is ${n\choose {n-1}}\,n^n$.
The second case is that we choose $n-2$ elements of $(x_1,x_2,\cdots,x_n)$ like $(x_1,x_2,\cdots,x_{n-2})$ and after that remove all possible cases that the $n$ places can be filled with the $(x_1,x_2,\cdots,x_{n-2},y)$ that numbers of the second case is ${n\choose {n-2}}\,{(n-1)}^n$.
The last two cases are, we choose $1$ elements of $(x_1,x_2,\cdots,x_n)$ like $(x_1)$ and after that remove all possible cases that the $n$ places can be filled with the $(x_1,y)$ that numbers of the second case is ${n\choose {1}}\,{(2)}^n$ and the last case is $(y,y,\cdots,y)$. 
Now, if we write our discussion by Inclusion–exclusion principle then we have:
$$
\begin{array}{ll}
n!=(n+1)^n-{n\choose {n-1}}\,{(n)}^n+{n\choose {n-2}}\,{(n-1)}^n
-\cdots+(-1)^{n-1}{n\choose {1}}\,{(2)}^n+(-1)^n \\
\\
n!=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}(k+1)^n
\end{array}
$$
Example:  Let we want to count permutation of numbers $(1,2)$ then we choose a number like $5$ and consider the set $(1,2,5)$. All possible cases that $(1,2,5)$ can fill $2$ places are $9$ cases but we remove cases $(1,5)$ and $(2,5)$ and because we remove two times the case $(5,5)$, we add one to our computation. 
I hope you fine it useful.
