Showing that $a^5$ and $a$ have the same last digit using Euler's Theorem. To show that, in base 10, $a^5$ and $a$ have the same final digit it is sufficient to prove that
$$a^5 \equiv a \bmod 10$$
Provided that $a$ and $10$ are coprime then Euler's Theorem proves the result:
$$a^{\phi(10)} \equiv 1 \bmod 10$$
where $\phi(10) = \phi(2\cdot 5) = \phi(2)\cdot \phi(5) = 1\cdot 4 = 4$, and hence
$$a^4 \equiv 1 \bmod 10 \implies a^5 \equiv a \bmod 10$$
The problem I have is: What if $a$ an $10$ are not coprime?
I have proven the result by going through all of the remainders one-by-one, e.g. if $a \equiv 5 \bmod 10$ then $a^5 \equiv 3125 \bmod 10 \equiv 5 \bmod 10$, etc.
I would like a more general proof, using Euler's Theorem if possible. Checking the remainders isn't practical for larger numbers, e.g. $\bmod 1000$.
 A: It's special case $\ p^i = 2,\ q^j = 5,\ \phi = 4,\ k=1\ $ of the following generalization of Euler $\!\rm\color{blue}{(E)}$.
Theorem $\ \ n^{\large k+\phi}\equiv n^{\large k}\pmod{p^i q^j}\ \ $ assuming that $ \ \color{#0a0}{\phi(p^i),\phi(q^j)\mid \phi},\, $  $\, i,j \le k,\,\ p\ne q.\ \ \ $ 
Proof $\ \ p\nmid n\,\Rightarrow\, {\rm mod\ }p^i\!:\  n^{ \phi}\equiv 1\,\Rightarrow\, n^{k + \phi}\equiv n^k,\, $ by $\  n^{\Large \color{#0a0}\phi} = (n^{\color{#0a0}{\Large \phi(p^{ i})}})^{\large \color{#0a0}m}\overset{\color{blue}{\rm (E)}}\equiv 1^{\large m}\equiv 1.$
$\qquad\quad\ \color{#c00}{p\mid n}\,\Rightarrow\, {\rm mod\ }p^i\!:\  n^k\equiv 0\,\equiv\, n^{k + \phi}\ $ by $\ n^k = n^{k-i} \color{#c00}n^i = n^{k-i} (\color{#c00}{p\ell})^i$ and $\,k\ge i.$
So $\ p^i\mid n^{k+\phi}\!-n^k.\,$ By symmetry $\,q^j$ divides it too, so their lcm $ = p^iq^j\,$ divides it too. $\ $ QED
Remark $\ $ Obviously the proof immediately extends to an arbitrary number of primes. This leads the way to Carmichael's Lambda function, a generalization of Euler's phi function.
A: What if $a$ and $10$ are not co-prime.
then $(2|a)$ or $(5|a)$ or $(10|a).$
$2$ divides $a$ and $10$ does not divide $a.$
Then we have these $4$ elements $\{2,4,6,8\}$
And these form a group.  It is cyclical of order $4.$
$a^4\equiv e \pmod{10}\\
a^5\equiv a\pmod{10}$
The identity element of this group turns out to be $6.$
$5$ divides $a$ and $10$ does not divide $a.$
$\forall n, a^n \equiv 5 \pmod {10}$
$10$ divides $a$
$\forall n, a^n \equiv 0 \pmod {10}$
More generally, for any modulo, the integers will fall into to cyclical groups, and the order of those groups will divide $\phi$
