Let $a, b$ in a group. If $ord(a) = n, ord(b) = m$, then $ab^{nm} = a^{nm}b^{nm} = e$ granted $ab = ba.$
I think $ab^{nm} = a^{nm}b^{nm} = e^mb^{nm} = b^{mn} = e^n = e.$ What does $ab = ba$ have to do with it? Do we need it to be able to say $b^{nm} = b^{mn}?$
If $a$ and $b$ don't commute, it's false that $(ab)^k=a^k b^k$ ; actually the case $k=2$ states precisely that $a$ and $b$ commute.
On the other hand, $b^{nm}=b^{mn}$ is always true (because $b$ commutes with itself, in a way).