Show that if $x\equiv 1 \pmod {m^k}, $then $x^m \equiv 1\pmod{m^{k+1}}$.

Question

Let $k\ge 1, m\ge 1.$ Show that if $x\equiv 1 \pmod {m^k}, $then $x^m \equiv 1\pmod{m^{k+1}}$.

First I noticed that the assumption would imply $x^m \equiv 1 \pmod{m^k}$, but that doesn't seem to work out. Then I tried:

$x\equiv 1 \pmod {m^k}$ implies that $x=m^k n+1$ for some $n$. If I raise it to the $m$-th power, I would get: $x^m=(m^k n+1)^m=\sum_{j=0}^m C(m,j)m^{kj}n^j$.

Now if the proposition is true, then $m^{k+1}$ must divide $1+\sum_{j=0}^m C(m,j)m^{kj}n^j$. Each term in the binomial expression are divisible by $m^{k+1}$ except for when $j=0$ and $j=1$. Take them out, we get $1+C(m,0)+C(m,1)m^kn=2+m^{k+1}n$. Everything look perfect except for that $2$...

Answer 1

Hint. If $$x=1+nm^k$$ then by the binomial theorem $$x^m=(1+nm^k)^m=1+\binom m1nm^k+\binom m2(nm^k)^2+\cdots\ .$$ Do a bit of simplification and see what you notice about every term (except the first) on the RHS.

In your solution, near the end, you have a $+$ sign instead of a $-$.