How to simplify $\frac{n^n - 1}{n - 1} \mod (n + 1)$? My question is how to simplify the following:
$$\frac{n^n - 1}{n - 1} \mod (n + 1)$$
I've tried a bunch of tricks (splitting into even and odd $n$ and using a difference of squares) but I can't seem to find anything that gives a clean answer.
Is there even a simplification for this?
EDIT: Apologies, I had $+$ instead of $-$
 A: You have changed your question, now it is easy.
Note that $n^n-1$ divides $n^2-1$ whenever $2 | n$. Hence, if $n$ is even, the answer is $0$, since $\frac{n^n-1}{(n+1)(n-1)}$ will be an integer.
Suppose that $n$ is odd. Note that $n \equiv -1 \mod n+1$. Hence, it is legitimate to replace $n$ by $-1$ in the modular expression, and this gives 
$$
\frac{n^n-1}{n-1} = \sum_{i=0}^{n-1} n^i \equiv \sum_{i=0}^{n-1} (-1)^i \equiv 1 \mod n+1
$$
because one factor of $1$ gets left out as $n$ is odd, so $n-1$ is even.
Hence, the answer is zero for even $n$, and $1$ for odd $n$.
Edit: You can multiply and check that $$
(n-1)  (1 + n + n^2 + \ldots + n^{n-1}) = n^n-1
$$
A: it's the same as the equivalance class of $n^n+1$ by definition of modulars
A: The left-hand side is the sum of powers of $n$.  Each power $n^a$ is equivalent to $(-1)^a\pmod{n+1}$
A: $\frac {n^n-1}{n-1} = \sum_\limits{i=0}^{n-1} n^i\\
n \equiv -1\pmod{n+1}\\
n^i \equiv -1^i\pmod{n+1}\\
\frac {n^n-1}{n-1} \equiv \sum_\limits{i=0}^{n-1} (-1)^i\pmod{n+1}$
$0$ when $n$ is even.
$1$ when $n$ is odd.
