New, elegant proofs for $\varphi(p^{k})=p^{k}-p^{k-1}$

A beautiful proof for the second identity you mention is probabilistic :

Take the set $\{1,...,n\}$ with the uniform probability measure (i.e. $P(\{i\}) = 1/n$ for any $i\in \{1,...,n\}$).

Then $\phi(n)/n = P(\{k, gcd(n,k) = 1\}) = P(\{k, \forall p$, prime, $(p$ divides $n) \implies (p$ doesn't divide $k)\} = P(\displaystyle\bigcap_{p\mid n} \{k, p$ doesn't divide $k\})$. Now it is easily proved that the events $(\{k, p$ divides $k\})_{p\mid n}$ are independent (just compute it) and so their complements are as well, which shows that

$\phi(n)/n = \prod_{p\mid n}(1- 1/p)$

That lets you conclude

Hint $\,\ \gcd(a,p^k)>1 \iff p\mid a \iff a\, \equiv\,\overbrace{ 1p,\,2p,3p,\ldots,\color{#c00}{p^{k-1}}p}^{\large\quad\color{#c00}{p^{\Large k-1}}\ \rm elements}\,\pmod{\!p^k}$

Thus there are $\,\color{#c00}{p^{k-1}}$ non-coprime residues, so $\,p^k - \color{#c00}{p^{k-1}}$ coprime residues mod $p^k$.

Yes, a one-line proof: $$(\mathbf Z/p^k\mathbf Z)^\times= \mathbf Z/p^k\mathbf Z\smallsetminus (p\mathbf Z/p^k\mathbf Z),\enspace\text{and}\quad p\mathbf Z/p^k\mathbf Z\simeq\mathbf Z/p^{k-1}\mathbf Z. $$ And a detail on a second line (well, a sesquiline…) for the isomorphism: \begin{align}p\mathbf Z/p^k\mathbf Z&\longrightarrow\mathbf Z/p^{k-1}\mathbf ,\\ px+p^k\mathbf Z&\longmapsto x+p^{k-1}\mathbf Z. \end{align}