#14. Using $(ab)^N=a^Nb^N$ and $(ab)^{N+1}=a^{N+1}b^{N+1}$, we get
\[
(ab)a^Nb^N=a^{N+1}b^{N+1}\Longrightarrow ba^N=a^Nb.
\]
Then we get $a^Nb^N=a^Nb\cdots b=ba^Nb\cdots b=\ldots=b^Na^N$, hence $(ab)^N=(ba)^N$. This is true for $N=2005,2006$, so $ab=ba$.
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