How Nikita Decrypts a Message

When Nikita receives an $E(m_i)$, she finds $m_i$ as follows:

$$m_i = E^{-1}(E(m_i)) = E(m_i)^d = (m_i^{e})^{d}=m_i.$$

The following proposition proves that the last equality holds.

Proposition 1.1   Let $n$ be a square-free integer and let $d,e\in\mathbb{N}$ such that $p-1\mid de-1$ for each prime $p\mid n$. Then $a^{de} \equiv a\pmod{n}$ for all $a\in\mathbb{Z}$.

Proof. Since $n\mid a^{de}-a$ if and only if $p\mid a^{de}-a$ for each prime divisor of $p$, it suffices to prove that $a^{de}\equiv a\pmod{p}$ for each prime divisor $p$ of $n$. If $\gcd(a,p)\neq 0$, then $a\equiv 0\pmod{p}$, so $a^{de}\equiv a\pmod{p}$. If $\gcd(a,p)=1$, then Fermat's Little Theorem asserts that $a^{p-1}\equiv 1\pmod{p}$. Since $p-1\mid de-1$, we have $a^{de-1} \equiv 1\pmod{p}$ as well. Multiplying both sides by $a$ shows that $a^{de}\equiv a\pmod{p}$. $\qedsymbol$