Let $d,n\in\mathbb{Z}^+$. The question is to prove that $d\mid n$ $\implies$ $\varphi(d)\mid\varphi(n)$. Here, $\varphi:\mathbb{Z}^+\to\mathbb{Z}^+$ is the Euler totient function.  For $n=1$, we must have $d=1$. Then, $\varphi(d)=\varphi(n)=1$ and thus $\varphi(d)\mid\varphi(n)$. For $n&gt;1$ and $d=1$, $\varphi(d)\mid\varphi(n)$ since $1\mid z$ for all $z\in\mathbb{Z}^+$. For $n,d&gt;1$, since $d\mid n$, we can express $\varphi(d)$ and $\varphi(n)$ as follows: $$\varphi(d)=p_1^{\alpha_1-1}(p_1-1)p_2^{\alpha_2-1}(p_2-1)...p_s^{\alpha_s-1}(p_s-1)$$ $$\varphi(n)=p_1^{\beta_1-1}(p_1-1)p_2^{\beta_2-1}(p_2-1)...p_s^{\beta_s-1}(p_s-1)q_1^{\gamma_1-1}(q_1-1)q_2^{\gamma_2-1}(q_2-1)...q_t^{\gamma_t-1}(q_t-1)$$ where  $p_i,q_i$ are distinct primes; $s,\alpha_i,\beta_i\in\mathbb{Z}^+$ with $\alpha_i\le\beta_i$; $q_i$ account for those prime factor(s) of $n$ which do not divide $d$. If such factors exist, then $t,\gamma_i\in\mathbb{Z}^+$, otherwise we don't have those terms in $\varphi(n)$.  Now, let $\omega=p_1^{\beta_1-\alpha_1}p_2^{\beta_2-\alpha_2}...p_s^{\beta_s-\alpha_s}q_1^{\gamma_1-1}(q_1-1)q_2^{\gamma_2-1}(q_2-1)...q_t^{\gamma_t-1}(q_t-1)$. Clearly $\omega\in\mathbb{Z}$. We then have $\varphi(d)\omega=\varphi(n)$, that is, $\varphi(d)\mid\varphi(n)$.  May I ask if my proof is valid? Many thanks!
Prove that if $d$ divides $n$ then $φ(d)$ divide $φ(n)$ for $φ$ denotes Euler’s $φ$-function.  I know that $d|n$ mean there exists some integer $k$ such that $n=kd$, but how can I use this to prove $φ(d)$ divide $φ(n)$
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