# Show that for $k \gt 0$ and $m \ge 1$, $x \equiv 1 \pmod {m^k}$ implies $x^m \equiv 1 \pmod {m^{k+1}}$.

Here's what I've done. I don't think I going the right way.

If $x \equiv 1 \pmod {m^k}$, then $x^m \equiv 1^m \equiv 1 \pmod{m^k}$.

$\Rightarrow x^m = 1^m + m^kn$, for some $n \in \Bbb Z$.

$\Rightarrow mx^m = m + m^{k+1}n$.

$\Rightarrow mx^m - m + 1 = 1 + m^{k+1}n$.

$\Rightarrow m(x^m - 1) + 1 = 1 + m^{k+1}n$.

...And I'm lost. Doesn't look like this is going anywhere.

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Suppose that $x\equiv 1 \pmod{m^k}$. Then there exists $n\in\mathbb{Z}$ such that $$x = 1 + nm^k$$ Taking the previous equation to the $m$th power, we have $$x^m = (1+nm^k)^m = \sum_{r=0}^m\binom{m}{r}n^rm^{kr}$$ Where the latter expansion is the binomial theorem. What happens to the terms of the summation when you take the equation mod $m^{k+1}$?
I'm assuming the terms cancel leaving $m^{k+1} + 1$? Not quite understanding. –  Tejen Shrestha Feb 21 '13 at 0:46
Well, sort of. What is the largest power of $m$ dividing each term of the sum if $r \ge 2$? Then what about $r=1$ and $r=0$? –  EuYu Feb 21 '13 at 0:49