Past exam of the options II course of the University of Oxford / 2005 / C / 2 / a

Euler's totient function $\varphi (n)$ counts the integers $1\le a\le n$ coprime to $n$, equivalently the invertible residue classes modulo $n$. Among $1,\dots ,p^e$, exactly the $p^{e-1}$ multiples of $p$ are not coprime to the prime power. HenceFactor $n={\prod }_p{p}^{e_p}$. Its distinct prime-power factors are pairwise coprime, so the permitted multiplicativity givesFor $n=1$, the empty product is one and agrees with $\varphi (1)=1$.

Use prime factorization and multiplicativity to check the identity one prime at a time. Fix $p$ and put $a=v_p(m)$, $b=v_p(n)$, $r=\min (a,b)$. The local factors to compare areIf either exponent is zero, $r=0$ and the equality follows from $\varphi (1)=1$. If both are positive, the prime-power formula makes both sidesMultiplying the local identities over all primes and using $\gcd (m,n)={\prod }_p{p}^{\min (v_p(m),v_p(n))}$ givesThis accounts for repeated prime factors shared by the two inputs, which ordinary coprime multiplicativity alone would not remove.