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 $(c,n) = 1$.
Since gcd$(b, 3) = 1$, $b^2 \equiv 1$ (mod $3$) by the Fermat's little theorem.
Since $b^{560} = (b^2)^{280}$, $b^{560} \equiv 1$ (mod $3$)
Best Answer
If $d | n$ then there is an $r \in \mathbb{Z}$ such that $dr \equiv 0 \mod n$ and $n$ does not divide $r$.
Suppose, by contradiction, there was any integer $a$ such that $ad = 1$. Then $1*r \equiv adr \equiv a*0 \equiv 0 \mod n$ so that $n|r$.
This shows that $d$ cannot have a multiplicative inverse modulo $n$. Thus $d^{n-1} \equiv 1 \mod n$ is impossible since this would imply that $d^{n-2}$ is a multiplicative inverse of $d$ modulo $n$.
The point is that zero divisors cannot have inverses.
Semi-related to your question is this interesting phenomenon.