Modular arithmetic. $x \equiv 1 \pmod {m^k}$ implies $x^m \equiv 1 \pmod {m^{k+1}}$ Show that for $k \gt 0$ and $m \ge 1$, $x \equiv 1 \pmod {m^k}$ implies $x^m \equiv 1 \pmod {m^{k+1}}$
This question has already been asked in SE (Show that for $k \gt 0$ and $m \ge 1$, $x \equiv 1 \pmod {m^k}$ implies $x^m \equiv 1 \pmod {m^{k+1}}$.), but I think it´s not really answered. ( the hint given was enough but I didn’t realize it before)
I tried with Fermat´s little theorem but I get nowhere.
note: If $a \equiv b \pmod m$, then $a \cdot t \equiv b \cdot t \pmod {m \cdot t}$ with $t \gt 0$ (don´t know if this is useful)
Any help would be appreciated. Thanks.
 A: $x \equiv 1 \pmod {m^k}\implies$ 
There is an integer $M$ so that $x = 1 + Mm^k$.
$x = (1 + M*m^k)$ and $x^m = (1+M*m^k)^m$ and by binomial theorem will equal $1 + Mm^{k+1} + $ bunch of terms all times $m$ to power greater than $k+1$.
i.e.
So $x^m = (1 + Mm^k)^m = 1 + m*M*m^k + \sum_{j=2}^m {m\choose j}M^jm^{jk}=$
$1 + M*m^{k+1} + m^{k+1}\sum+{j=2}^m {m\choose j}M^jm^{jk-(k+1)}$.  
(Note: If $j \ge 2$ then $jk-(k+1)=(j-1)k - 1\ge k-1 \ge 0$ so )
$Mm^{k+1} \equiv 0 \pmod{m^{k+1}}$ and $m^{k+1}\sum+{j=2}^m {m\choose j}M^jm^{jk-(k+1)}\equiv 0 \pmod{m^{k+1}}$
$x^m \equiv 1\pmod {m^{k+1}}$.
A: We have $x^m - 1 = (x-1) (x^{m-1} + x^{m-2} + \cdots + 1)$.  Now, by hypothesis, $m^k \mid x-1$.  On the other hand, since $k > 0$, we also have $x \equiv 1 \pmod{m}$, so $x^{m-1} \equiv 1 \pmod{m}$, $x^{m-2} \equiv 1 \pmod{m}$, ..., $1 \equiv 1 \pmod{m}$.  Therefore, $x^{m-1} + x^{m-2} + \cdots + 1 \equiv 1 + 1 + \cdots + 1 = m \equiv 0 \pmod{m}$, so $m \mid x^{m-1} + x^{m-2} + \cdots + 1$.
A: Consequence of binomial expansion. If $$x=m^kn+1$$ then $$x^m=(m^kn+1)^m$$ the expansion of this has the exponent on $m$ in each term, except the last 2, greater than k( in fact multiples of k prior to coefficients). The last term is $1^m$ and the second last has ${{m}\choose {m-1}}=m$  as a coefficient. fleabloods answer tipped me off to my mistakes. 
