For all odd prime $p$ and integer $n\geq 2$, prove that $(1+ap)^{p^{n-2}} = 1+ap^{n-1}$ modulo $p^n$.
This is the fact that is commonly used to show that the unit group of the ring $\mathbb Z/p^n\mathbb Z$ is cyclic for $n>2$ assuming the existence of generators of $(\mathbb Z/p^2\mathbb Z)^\times$. But I have only seen arguments using induction on $n$. Is there a more elegant proof that one can actually observe and come up with this congruence equation? Thank you.
Lemma: $p^{m-k}\mid \dbinom{p^m}{k}$ when $1\le k\le m$
proof: $\dbinom{p^m}{k} = \dfrac{p^m\cdot(p^m-1)\cdots (p^m-k+1)}{k!}=\dfrac{p^m\cdot\color{blue}{(p^m-1)\cdots (p^m-k+1)}}{k\cdot \color{blue}{(k-1)!}}$
Since $\color{blue}{(p^m-1)\cdots (p^m-k+1)}$ is a product of $k-1$ consecutive integers, it is divisible by $\color{blue}{(k-1)!}$
Also since $k\lt p^k$, the highest exponent of $p$ in $k$ is $\le k$.
Thus $p^{m-k}$ is a factor of $\dbinom{p^m}{k}$
$\square$
$(1+ap)^{p^{n-2}} = 1+\dbinom{p^{n-2}}{1}(ap)^1 + p^n(\cdots) \equiv 1+ap^{n-1}\pmod{p^n}$