Prove that $37$ is a factor of $(10^{2014} + 10^{2015} + 2)^{2016} − 1$ How can I prove that $37$ is a factor of $(10^{2014} + 10^{2015} + 2)^{2016} − 1$, probably using basic number theory?

Below is a somewhat not so elementary approach:
Notice $2016=36\times56$, and by Fermat's little theorem, for any $37\not\mid a$, we have $a^{36}\equiv1\pmod{37}$.
So $(10^{2014} + 10^{2015} + 2)^{2016} − 1\equiv-1\text{ or } 0\pmod{37}$.
But we know $10^{2016}\equiv1\pmod{37}$ so $10^{2015}\equiv10^{-1}\equiv2^{-1}\times5^{-1}\equiv19\times15\equiv26\pmod{37}$ and $10^{2015}+2\equiv28\pmod{37}$.
Moreover we see that if $10^{2014} + 10^{2015} + 2\equiv10^{2014}+28\equiv0\pmod{37}$, then $10^{2014}\equiv9\pmod{37}$.
However this is impossible: $10^1\equiv10\pmod{37}, 10^{2}\equiv26\pmod{37}, 10^{3}\equiv1\pmod{37}$.
Hence $10^{2014} + 10^{2015} + 2\not\equiv0\pmod{37}$ and $(10^{2014} + 10^{2015} + 2)^{2016}\equiv1\pmod{37}$.
Hope this helps.
