Totient function and Euler's Theorem Given $\big(m, n\big) = 1$, Prove that 
$$m^{\varphi(n)} + n^{\varphi(m)} \equiv  1 \pmod{mn}$$
I have tried saying 
$$\text{let }(a, mn) = 1$$
$$a^{\varphi(mn)} \equiv 1 \pmod{mn}$$
$$a^{\varphi(m)\varphi(n)} \equiv 1 \pmod{mn}$$
$$(a^{\varphi(m)})^{\varphi(n)} \equiv 1 \pmod{mn}$$
but I can't see where to go from here. I'm trying to somehow split the $a^{\varphi(mn)}$ into an addition so I can turn it into $m^{\varphi(n)} + n^{\varphi(m)} \equiv 1 \pmod{mn}$.

Hint What is $m^{\varphi(n)} + n^{\varphi(m)}  \pmod{m}$? What about $\pmod n$?
