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I want to prove the following:

Show that if $m$ is a positive integer, $m>1$, then $a^m \equiv a^{m-\phi(m)}\pmod m$ for all positive integers $a$.

I imagine the Fundamental Theorem of Arithmetic and Euler's Theorem will be useful here, but I don't quite see how.

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    In order to use Euler's totient theorem you need $a$ and $m$ to be coprime . – 2017-02-27
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    Indeed; this is why I suggested use of the FTA. – 2017-02-27
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    Let $p$ a prime dividing $m$. Write $m = p^k \cdot \tilde{m}$ with $p\nmid \tilde{m}$. You want to show that $a^m \equiv a^{m-\phi(m)} \pmod{p^k}$. That's clear from Euler's theorem if $p\nmid a$ since $\phi(p^k) \mid \phi(m)$. And if $p\mid a$ then $a^m \equiv 0 \pmod{p^k}$ since $k < m$, so you need to show $m - \phi(m) \geqslant k$. Any ideas for that? – 2017-02-27
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    http://math.stackexchange.com/questions/1942239/generalisation-of-eulers-theorem/1942869#1942869 – 2017-02-27

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