Q:

Explain why $a^k\equiv1 \pmod {m_{1}\cdot m_{2}}$ if...

Supppose that $\gcd(m_{1},m_{2})=1$ and that for some $a$ and $k\ge1$ we have that $a^k\equiv1 \pmod{m_{1}}$ and that $a^k\equiv1\pmod{m_{2}}$.  Explain why  $a^k\equiv1 \pmod{m_{1}\cdot m_{2}}$.

A:

Relabel $a^k = a$ because the $k$ is quite irrelevant actually.
(I think the OP is confused about Fermat's little and might want to study it again.)
Also relabel $m_1$ and $m_2$ as $b$ and $c$.
( I find that easier thats all )
We are given : $a = 1 $ $mod$ $ b$
and
$a = 1 $ $mod$ $ c$
and $gcd(b,c) = 1$.
Thus $a = b x + 1$ and $a= c y + 1$ for some integer $b$ and $c$.
It follows that $a-1 = b x = c y$.
Thus $a-1$ is both divisible by $b$ and $c$. 
$b,c | (a-1) => lcm(b,c) | (a-1).$
Also
$lcm(b,c) | (a-1)$ and $gcd(b,c) = 1$ => $lcm(b,c) = bc$ ( because of unique prime factorization of any integer $> 1$ )
So $bc | (a-1).$
Hence $a-1 = b c z$ for some integer $z$ because $b$ and $c$ have no common divisor.
Since $a-1 = b c z$ it must follow that $a = b c z + 1$ which is equivalent to $a = 1$ $ mod $ $bc$  QED

