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?

\begin{eqnarray}
\color{red}{10^3} &\color{red}{\equiv}& \color{red}{1\,\,\, mod \,37}\\
10^{2010}=(10^3)^{67} &\equiv& 1\,\,\, mod \,37\\
10^{2014}=10^{2010}\times10^3\times10 &\equiv& 10\,\,\, mod \,37\\
10^{2015}=10^{2010}\times10^3\times100 &\equiv& 100\,\,\, mod \,37\\
\end{eqnarray}
so
\begin{eqnarray}
10^{2014}+10^{2015}+2 &\equiv& 10+100+2\,\,\, mod \,37\\
10^{2014}+10^{2015}+2 &\equiv& 1\,\,\, mod \,37\\
\Big(10^{2014}+10^{2015}+2\Big)^{2016} &\equiv& 1\,\,\, mod \,37\\
\Big(10^{2014}+10^{2015}+2\Big)^{2016}-1 &\equiv& 0\,\,\, mod \,37
\end{eqnarray}
