<s>[INST] Proving in different ways that $n^{n-1}-1$ is divisible by $(n-1)^2$. I have this amazing exercise which explicitly says prove in at least six different way that 
$n^{n-1}-1$ is divisible by $(n-1)^2$ where $n$ is an integer.
So far I have only prove it as follows: By geometric sum we have
$$\sum_{k=0}^{n-2}n^k=\frac{n^{n-1}-1}{n-1}$$
However, since $n\equiv 1\mod (n-1)$ we have 
$$\frac{n^{n-1}-1}{n-1}=\sum_{k=0}^{n-2}n^k\equiv \sum_{k=0}^{n-2}1^k\mod(n-1)$$
that is $$\frac{n^{n-1}-1}{n-1}=\sum_{k=0}^{n-2}n^k\equiv n-1\mod(n-1)\Longleftrightarrow  \frac{n^{n-1}-1}{n-1} \equiv 0\mod(n-1)$$
and this prove that $(n-1)^2$ divides $n^{n-1}-1$.

Does anyone have another approach different from mine.?
  Note that I do necessary need one to give all six approaches as was asked in the exercise one or two is plainly enough for me.  

 [/INST] $$
\begin{align}
n^{n-1}-1 &= (n-1) ( n^{n-2} \color{red}{- 1} + n^{n-3} \color{red}{- 1} +\ldots + n^2 \color{red}{- 1} + n \color{red}{- 1} + 1 \color{red}{+n-2}) \\
 &= (n-1) \big(\,(n-1)(\ldots) + (n-1)(\ldots)+ \ldots + (n-1)(n+1)+ (n-1)+ (n-1))\,\big)
\end{align}
$$
</s>