Question 1207212
Unfortunately, that is not true, since taking n=2 gives that 10 is divisible by 4, which is clearly false.
Did you mean {{{(n+1)^n-1}}} instead? In that case, the statement is true, and the proof is as follows. 
To show this, we can expand out {{{(n+1)^n}}} with the binomial theorem. Doing so, we get {{{1+nC1*n+nC2*n^2+...}}}. Note that all the terms after {{{1}}} are divisible by {{{n^2}}} (the second term is just {{{n^2}}}), so we have {{{(n+1)^n}}} is congruent to 1 mod {{{n^2}}}, which means that {{{(n+1)^n-1}}} is congruent to 0 mod {{{n^2}}}, aka {{{(n+1)^n-1}}} is divisible by {{{n^2}}}.