Prove that if $d$ divides $n$ then $φ(d)$ divide $φ(n)$ for $φ$ denotes Euler’s φ-function. 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 $φ(n)$
It is true for $\rm\color{#0a0}{prime\ powers}$, so for all naturals, because  $\phi$ is $\color{#c00}{\rm multiplicative},$ i.e. $\,\phi(ab) = \phi(a)\phi(b)\,$ for coprime $a,b,\,$ therefore $\ \phi(p_1^{n_1}\cdots p_k^{n_k}) = \phi(p_1^{n_1})\cdots \phi(p_k^{n_k})\,$ for distinct primes $\,p_i$. Explicitly

$$\begin{eqnarray} &amp;&amp;\qquad\qquad\qquad\ \ \ p^i \cdots q^j\ \mid\  p^I\cdots q^J\quad\ {\rm for\ distinct\ primes}\ \ p,\ldots, q \\ 
\Rightarrow\ &amp;&amp;\qquad \quad\ \ \color{#0a0}{p^i\mid p^I},\qquad\quad\ \ldots,\qquad\quad \ \  q^j\mid q^J \\ 
\Rightarrow\ &amp;&amp;\color{#0a0}{(p\!-\!1)p^{i-1}\mid (p\!-\!1)p^{I-1}},\ldots,(q\!-\!1)q^{j-1}\mid (q\!-\!1)q^{J-1} \\
\Rightarrow\ &amp;&amp;\qquad\ \color{#0a0}{\phi(p^i)\mid \phi(p^I)},\qquad\ldots,\qquad\,\phi(q^j)\mid\phi(q^J)\\
\Rightarrow\  &amp;&amp;\qquad\qquad\ \, \phi(p^i)\cdots\phi(q^j)\mid \phi(p^I)\cdots\phi(q^J)\\
\Rightarrow\  &amp;&amp;\qquad\qquad\qquad\!\phi(p^i\cdots q^j)\mid \phi(p^I\cdots q^J)\quad\ \   \rm by\ \phi\ is\ \color{#c00}{multiplicative}
\end{eqnarray}$$

Remark $\,\ $ Similarly, generally a "multiplicative" statement about a multiplicative function is true if it is true for prime powers.
You may know that $\phi(pd)=p\phi(d)$ and $\phi(pd)=(p-1)\phi(d)$ if $p\not\mid d$ (where $p$ is a prime).