Help using the inclusion-exclusion principle to prove $\sum_{n|d}\mu(d)\frac{n}{d}=\varphi(n)$

Help using the inclusion-exclusion principle to prove
$\sum_{n|d}\mu(d)\frac{n}{d}=\varphi(n)$

$(1)$Using the Dirichlet convolution theorem,
$$\sum_{n|d}\mu(d)\frac{n}{d}=\sum_{ab=n}\mu(a)b$$ Let $n=p_1\cdot
p_2\cdots p_n\cdot k$, where $k$'s prime factorisation contains $p_k$ only
if $ k\in\{1,2,...,n\}$. Note that, for nonzero $\mu(a)$, $k|b$, else $a$
wouldn't be square-free. $(2)$ Thus there is a bijection between all $a$
such that $k|b$ and all subsets ($S$) of $\{p_1,p_2,...,p_n\}$. If $|S|$
is odd, $\mu(a)=-1$, if even, $\mu(a)=1$. Hence the original sum is equal
to $$\displaystyle k\left(1+\sum_{\begin{matrix} 1\le a,b \le n\\ a<b
\end{matrix} }p_ap_b+\sum_{\begin{matrix} 1\le a,b,c,d \le n\\ a<b<c<d
\end{matrix}}p_ap_bp_cp_d+\cdots\right)$$ $$\displaystyle -k\left(\sum_{
1\le a \le n}p_a+\sum_{\begin{matrix} 1\le a,b,c \le n\\ a<b<c
\end{matrix} }p_ap_bp_c+\cdots\right)$$ $$=\displaystyle k\left(1-\sum_{
1\le a \le n}p_a+\sum_{\begin{matrix} 1\le a,b \le n\\ a<b \end{matrix}
}p_ap_b-\sum_{\begin{matrix} 1\le a,b,c \le n\\ a<b<c \end{matrix}
}p_ap_bp_c+\sum_{\begin{matrix} 1\le a,b,c,d \le n\\ a<b<c<d
\end{matrix}}p_ap_bp_cp_d-\cdots\right)$$
$$=\displaystyle n\left(\frac{1}{p_1\cdots p_n}-\sum_{ 1\le a \le
n}\frac{p_a}{p_1\cdots p_n}+\sum_{\begin{matrix} 1\le a,b \le n\\ a<b
\end{matrix} }\frac{p_ap_b}{p_1\cdots p_n}-\sum_{\begin{matrix} 1\le a,b,c
\le n\\ a<b<c \end{matrix} }\frac{p_ap_bp_c}{p_1\cdots
p_n}+\cdots\right)$$
This looks quite a lot like the inclusion-exclusion principle might work
well here to show that $\sum_{ab=n}\mu(a)b=\varphi(n)$, but I'm not well
versed enough in the principle to do so. Any help? Algebraically, it's
possible to prove the initial theorem algebraically using
$$\displaystyle\left(\frac{1}{p_1\cdots p_n}-\sum_{ 1\le a \le
n}\frac{p_a}{p_1\cdots p_n}+\sum_{\begin{matrix} 1\le a,b \le n\\ a<b
\end{matrix} }\frac{p_ap_b}{p_1\cdots p_n}-\sum_{\begin{matrix} 1\le a,b,c
\le n\\ a<b<c \end{matrix} }\frac{p_ap_bp_c}{p_1\cdots
p_n}+\cdots\right)=\prod_{k=1}^n\left(1-\frac{1}{p_k}\right), $$
but it would be nice to have an inclusion-exclusion (I think this is
combinatorial) argument.