Proving that $x^a=x^{a\,\bmod\,{\phi(m)}} \pmod m$ i want to prove $x^a \equiv x^{a\,\bmod\,8} \pmod{15}$.....(1)
my logic:
here, since $\mathrm{gcd}(x,15)=1$, and $15$ has prime factors $3$ and $5$ (given) we can apply Euler's theorem.
we know that $a= rem + 8q$, where $8= \phi(15)$,
$x^a \equiv x^{rem}. (x^8)^q \pmod{15}$......(2)
applying Euler's theorem we get:
$x^a \equiv x^{rem} \pmod{15}$......(3)
Is this proof correct or should I end up in getting $x^a \equiv x^a \pmod {15}$...(4)

If $b\equiv a\pmod m, b$ will be equal to $a\iff 0\le a<m$ 
For example, $m-2\equiv m-2\pmod m, 13\equiv 13\pmod {15}$
but, $m+2\equiv 2\pmod m, 17\equiv 2\pmod {15}$

If $b\equiv c\pmod{\phi(m)} ,$i.e., if $b=c+d\phi(m)$
$y^b=y^{c+d\phi(m)}=y^c\cdot(y^{\phi(m)})^d\equiv y^c \pmod m$ for $(y,m)=1$
Here $\phi(15)=\phi(3)\phi(5)=2\cdot4=8$
Observe that this condition is no way necessary as proved below.
If $y^b\equiv y^d\pmod m$ where $b\ge c$
$y^{b-d}\equiv1\pmod m\iff ord_my\mid (b-d)$ does not need to divide $\phi(m)$ unless $ord_my=\phi(m)$ where $y$ is a primitive root of $m$.
