$\varphi(360)=96$ by both inclusion-exclusion and the prime-power product formula

Given: The positive integer $360$.

Its canonical prime factorisation is $360=2^3\cdot3^2\cdot5$ (For $n \ge 1$ and any injective list $p : r \to \mathbb{Z}$ of primes containing every prime divisor of $n$, one has $n = \prod_{i<r} p_i^{\,v_{p_i}(n)}$; the exponents are determined by $n$, and $v_q(n) = 0$ for every prime $q$ outside the list).

Euler's formula gives $\varphi(n)=\prod_{i<r}\bigl(p_i^{k_i}-p_i^{k_i-1}\bigr)$ over the distinct prime divisors of $n$ and, after carrying the natural numbers into $\mathbb{Q}$, the equivalent form $\varphi(n)=n\prod_{i<r}\left(1-\frac1{p_i}\right)$ (Euler's product formula $\varphi(n)=n\prod_{p\mid n}(1-1/p)=\prod_{p^k\parallel n}(p^k-p^{k-1})$ for $n\ge1$, stated through a finite injective list of its prime divisors). Multiplying out that second product over the subsets of $\{p_0,\ldots,p_{r-1}\}$ is the inclusion-exclusion display used below; the cited theorem states the two products, not that display.