proof that the sum of the iterated totient function is always odd

Given a positive integer $n$, it is always the case that

$$2|̸\sum _{i=1}^{c+1}{\varphi }^i(n),$$

where ${\varphi }^i(x)$ is the iterated totient function and $c$ is the integer such that ${\varphi }^c(n)=2$.

Accepting as proven that $n>\varphi (n)$ and $2|\varphi (n)$ for $n>2$, it is clear that summing up the iterates of the totient function up to $c$ is summing up a series of even numbers in descending order and that this sum is therefore itself even. Then, when we add the $c+1$ iterate, the sum turns odd.

As a bonus, this proves that no even number can be a perfect totient number.