Easiest proof for $\sum_{d|n}\phi(d)=n$ To prove $\sum_{d|n}\phi(d)=n$. What is the easiest proof for this to tell my first year undergraduate junior. I do not want any Mobius inversion etc only elementry proof. Tthanks!

Write $n = \prod p^{a_p}$
The divisors of $p$ are
$$
\prod p^{b_p}, b_p \le a_p
$$
so the sum on the left hand side is
$$
\sum_{b_2=0}^{a_2}\dots \sum_{b_P=0}^{a_P} 
      \phi(\prod_{p \text{ prime}, p|n, =2}^P p^{b_p})
= \sum_{b_2=0}^{a_2}\dots \sum_{b_P=0}^{a_P} 
\prod_{p \text{ prime}, p|n, b_p>0} \left(1-\frac 1p\right) p^{b_p}
\\  = \prod_{p \text{ prime}, p|n, =2}^P \left[1 + 
\sum_{b_p = 1}^{a_p} \left(1-\frac 1p\right) p^{b_p}
\right] = \prod_{p \text{ prime}, p|n, =2}^P p^{a_p} = n
$$
