Totient divisibility along divisors
= Totient divisibility along divisors
{title2=$d\mid n\Longrightarrow\varphi(d)\mid\varphi(n)$}
Write $n=\prod p^{a_p}$ and $d=\prod p^{b_p}$, with $0\le b_p\le a_p$. The prime-power formula for the <Euler totient function> gives an integer contribution to $\varphi(n)/\varphi(d)$ from each prime: it is $p^{a_p-b_p}$ when $b_p>0$, and $p^{a_p-1}(p-1)$ when $b_p=0<a_p$. Multiplying proves the <divisibility> assertion, including $d=1$ with $\varphi(1)=1$.