GCD via Euler's totient

Partition. For each divisor $d\mid n$, let $S_d=\{k\in\{1,\ldots,n\}:\gcd(k,n)=d\}$.

Size. $k\in S_d$ iff $k=dm$ with $1\le m\le n/d$ and $\gcd(m,n/d)=1$. So $|S_d|=\varphi(n/d)$.

Sum. The sets $S_d$ partition $\{1,\ldots,n\}$, so $\sum_{d\mid n}\varphi(n/d)=n$, i.e. $\sum_{d\mid n}\varphi(d)=n$.

Conclusion. The identity holds.

Example. $n=6$: $\varphi(1)+\varphi(2)+\varphi(3)+\varphi(6)=1+1+2+2=6$. ✓