Proof of the Euler Generalisation of Fermat's Little Theorem using modular arithmetic

Here is another way to prove Euler's generalization. You do not need to know the formula of $\varphi(n)$ for this method which I think makes this method elegant.

Consider the set of all numbers less than $n$ and relatively prime to it. Let $\{a_1,a_2,...,a_{\varphi(n)}\}$ be this set.

Consider a number $c < n$ and relatively prime to it i.e. $c \in \{a_1,a_2,\ldots,a_{\varphi(n)}\}$.

First observe that for any $a_i$, $c a_{i} \equiv a_{j} \pmod{n}$ for some $j$. (True since $c$ and $a_i$ are themselves relatively prime to $n$, their product has to be relatively prime to $n$).

And if $c a_{i} \equiv c a_{j} \pmod{n}$ then $a_i = a_j$. (True as cancellation can be done since $c$ is relatively prime to $n$).

Hence, if we now consider the set $\{ca_1,ca_2,...,ca_{\varphi(n)}\}$ this is just a permutation of the set $\{a_1,a_2,...,a_{\varphi(n)}\}$.

Thereby, we have $\displaystyle \prod_{k=1}^{\varphi(n)} ca_k \equiv \prod_{k=1}^{\varphi(n)} a_k \pmod{n}$.

Hence, we get $\displaystyle c^{\varphi(n)} \prod_{k=1}^{\varphi(n)} a_k \equiv \prod_{k=1}^{\varphi(n)} a_k \pmod{n}$.

Now, note that $\displaystyle \prod_{k=1}^{\varphi(n)} a_k$ is relatively prime to $n$ and hence you can cancel them on both sides to get

$$c^{\varphi(n)} \equiv 1 \pmod{n}$$ whenever $(c,n) = 1$.

Part 2 can be proven in many different ways; what your lecturer suggests, if you already know the identity, will do it. Just notice that if $p|rs$, then either $p|r$ or $p|s$, and not both, since $\gcd(r,s)=1$. So we have: $$\varphi(rs) = rs\prod_{p|rs}\left(1 - \frac{1}{p}\right) = rs\left(\prod_{p|r}\left(1 - \frac{1}{p}\right)\right)\left(\prod_{p|s}\left(1-\frac{1}{p}\right)\right),$$ and go from there.

One you have this, the proof of the Fermat-Euler Theorem reduces to proving that if $\gcd(a,p)=1$, then $a^{\varphi(p^n)} \equiv 1 \pmod{p^n}$ for every positive integer $n$; you can use 2 and the Chinese Remainder Theorem to prove that this suffices, and then you can use 1 to try to establish the condition modulo a prime power.