RSA finding the inverse of the public exponent
The answer comes from Euler's Theorem. Note: math below is done modulo $N$ unless otherwise specified and draws heavily from group theory.

That theorem says that any element of a group (say $m$) raised to the order of the group, in this case $\phi(N)$ is congruent to $1$ (i.e.,  $m^{\phi(N)}\equiv 1\bmod{N}$). Furthermore, this holds for multiples of $\phi(N)$ (i.e., $c\cdot\phi(N)$ where c is an integer).

In the case of RSA, you find $e,d$ such that $ed\equiv 1\bmod{\phi(N)}$. Note then that $ed=c\cdot\phi(N)+1$, where $c$ is some integer.

Thus $c^{d}=m^{ed}=m^{c\cdot\phi(N) + 1}=m^{c\cdot\phi(N)}\cdot m^1$. From the Euler's Theorem, $m^{c\cdot\phi(N)}=1$, so that term goes away and you are left with the original message.