If $t^p \equiv 1$ (mod $p^{n+1}$), can we say $t \equiv 1$ (mod $p^n$)? If $t^p \equiv 1$ (mod $p^{n+1}$), then $t \equiv 1$ (mod $p^n$). Is this statement true?
My progress now is that suppose $t\equiv k $ (mod $p^n$), so $t=ap^n+k$ for some integer $a$, so $t^p\equiv k^p\equiv 1 $ (mod $p^{n+1}$). Therefore, the order of $k$ should be $p$. Are there any elements whose order is $p$ in this ring? If such element exists, then I think the statement is not true, but how can I change the conclusion of the statement to make it true? Thanks a lot!
 A: The theorem is true only if $p$ is an odd prime. You can prove it by induction on $n$.
Base case ($n=1$): $t^p \equiv 1 \pmod{p^2} \implies t^p \equiv 1 \pmod p$.
By Fermat's little theorem, this proves that $t \equiv 1 \pmod p$.
Induction step: Assume it's true for $n-1$.
$$ t^p \equiv 1 \pmod{p^{n+1}} \implies t^p \equiv 1 \pmod{p^n} $$
By induction, this implies $t \equiv 1 \pmod{p^{n-1}}$ and therefore,
$$ \begin{align} t & = 1 + kp^{n-1} \\ t^p & = (1 + kp^{n-1})^p \\ t^p & = 1+ kp^n + mp^{n+1} \end{align}$$
The last equality follows from the binomial theorem and the fact that $p \mid {p \choose k}$ for $1 \leq k \leq p-1$. The requirement that $t^p \equiv 1 \pmod{p^{n+1}}$ forces $p$ to divide $k$ which in turn proves that $t \equiv 1 \pmod{p^n}$.
Why does this fail for $p=2$? There's a hint given above.
The analogous theorem for $p=2$ is
$$t^2 \equiv 1 \pmod{2^{n+1}} \iff t \equiv \pm 1 \pmod {2^n}$$
(Hint: If you take care of a single value of $n$, the above theorem also follows.)
