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 $m^{\varphi(n)} + n^{\varphi(m)} \equiv 1 \pmod{mn}$.
On
Hint $\ $ If $\rm\,(m,n)=1\,$ then $\rm\ x \equiv a m^{\varphi(n)}\!\! + b n^{\varphi(m)}\,\ (mod\ mn)\iff \begin{eqnarray} x\equiv a&&\rm (mod\ n)\\ \rm x\equiv b&&\rm (mod\ m)\end{eqnarray}$
Yours is the special constant case of CRT: $\rm\:a = b\ (= 1),\:$ so $\rm\:mn\mid x\!-\!a\iff m,n\mid x\!-\!a$
Hint What is $m^{\varphi(n)} + n^{\varphi(m)} \pmod{m}$? What about $\pmod n$?