A proof involving the Euler phi function

Let,

$$d={p_{1}}^{k_{1}}{p_{2}}^{k_{2}} \cdot \cdot\cdot {p_{j}}^{k_{j}}$$

$\text{&}$

$$n={p_{1}}^{l_{1}}{p_{2}}^{l_{2}} \cdot \cdot\cdot {p_{j}}^{l_{j}}{p_{j+1}}^{l_{j+1}} \cdot \cdot \cdot {p_{r}}^{l_{r}} $$

Where $\{p_{i}\}_{i=1}^r$ are prime numbers and $(k,j,r) \in \mathbb W$

Since $ \ d|n \implies l_{i} \geq k_{i} \ \ \forall \ \ i \in \{1,2,3,\cdot\cdot\cdot,j\} $

From the definition of $\phi$ function, we have,

$$\phi(d)={p_{1}}^{(k_{1}-1)}{p_{2}}^{(k_{2}-1)} \cdot \cdot\cdot {p_{j}}^{(k_{j}-1)} \prod_{i=1}^{j} (p_{i}-1) $$

$\text{&}$

$$\phi(n)={p_{1}}^{(l_{1}-1)}{p_{2}}^{(l_{2}-1)} \cdot \cdot\cdot {p_{r}}^{(l_{r}-1)} \prod_{i=1}^{r} (p_{i}-1) $$

Since $l_{i} \geq k_{i} \implies l_{i} -1 \geq k_{i} -1 \ \ \forall \ \ i \in \{1,2,3,\cdot\cdot\cdot,j\} $. Also, $r \geq j$

$\implies \phi(d)|\phi(n)$

Q.E.D.