This follows from Abel's binomial theorem (see equation (5) here): $$ (x+y)^n = \sum_{i=0}^n \binom{n}{i} x(x-ai)^{i-1}(y+ai)^{i-1}. $$ If we take $y=n$ and $a=-1$, we get $$ (x+n)^n = \sum_{i=0}^n \binom{n}{i} x(x+i)^{i-1}(n-i)^{i-1}. $$ Now differentiate both sides with respect to $x$ and set $x=0$ to get the desired identity.

This follows from Abel's binomial theorem (see equation (5) here): $$ (x+y)^n = \sum_{i=0}^n \binom{n}{i} x(x-ai)^{i-1}(y+ai)^{n-i}. $$ If we take $y=n$ and $a=-1$, we get $$ (x+n)^n = \sum_{i=0}^n \binom{n}{i} x(x+i)^{i-1}(n-i)^{n-i}. $$ Now differentiate both sides with respect to $x$ and set $x=0$ to get the desired identity.