Euler's Theorem

Theorem
Let $a, m \in \Z: a \perp m$.

Let $\phi \left({m}\right)$ be the Euler Phi Function of $m$.

Then:
 * $a^{\phi \left({m}\right)} \equiv 1 \pmod m$

Proof
Since $a \perp m$, the residue class modulo $m$ of $a$ belongs to the Multiplicative Group of Integers Modulo m $\left({\Z'_m, \times}\right)$.

Let $ k = \left|{\left[\!\left[{a}\right]\!\right]_m}\right|$ where $\left[\!\left[{a}\right]\!\right]_m \in \Z'_m$.

By Order of Element Divides Order of Finite Group, $k \backslash \left|{\Z'_m}\right|$.

Now $\left|{\Z'_m}\right| = \phi \left({m}\right)$, from the definition of the Euler Phi Function.

Thus: