It is easy to show that the condition
(i) $(ab)^n=a^nb^n$ for all $a,b\in G$, for consecutive integers $n=N,N+1$
is equivalent to that
(ii) $(ab)^N=a^Nb^N$ and $ab^N=b^Na$ for all $a,b\in G$.
If $G$ is a non-abelian finite group of order $N$, then $x^N=e$ (the identity) for all $x\in G$, hence (ii) is true for all $a,b\in G$.
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