Show that $a^p \equiv 1$ (mod $p^n$) $\Rightarrow a \equiv 1$ (mod $p^{n-1}$) I'm trying to show that $a^p \equiv 1$ (mod $p^n$) $\Rightarrow a \equiv 1$ (mod $p^{n-1}$), where $p$ is an odd prime.

Since $a^p \equiv 1$ (mod $p^n$) $\iff aa^{p-1} \equiv 1$ (mod $p^{n-1}p$), I was hoping to use Fermat's theorem, which states that $a^{p-1} \equiv 1$ (mod $p$). I therefore tried substituting $a^{p-1}$ with $pk + 1$ in the equation but I didn't manage to solve it like that. 

I've been trying to solve it algebraically for some time but is no where close so I could use some help and ideas.
Let $a=1+p^iX$, where $X$ is coprime to $p$.

Then $a^p-1=\begin{pmatrix}p\\1\\\end{pmatrix}p^iX+ ... +p^{pi}X^p$, where all the coefficients on the RHS except possibly the last are divisible by $p$. 

We know that $p^{n} $ divides the LHS so $i&gt;0$. Then, on the RHS, the first term has the lowest power of $p$ and so its power, $i+1$, must be at least $n$.
If $a^p≡1 \mod (p^n)$  then $\phi (p^n)=p$ ⇒ $p^{n-1}(p-1)=p$ ⇒ $p^{n-2}(p-1)=\phi(p^{n-1})=1$ ⇒ $\phi(p^{n-1})=1$ ⇒ $a^{\phi(p^{n-1})}=a^1≡1 \mod (p^{n-1})$.