# Math Help - Congruence and Euler's function

1. ## Congruence and Euler's function

Show if $m>1$ then $a^m \equiv a^{m-\phi(m)}(mod \ m)$ for all natural numbers $a$.
Basic Euler function is this $a^{\phi(m)} \equiv 1 (mod \ m)$ but how to build it up in order to answer the above question?
2. $a^{\phi(m)} \equiv 1\implies a^{-\phi(m)}\equiv 1$. Now multiply both sides by $a^m$. (Here negative exponents means inverses mod $m$).