How is this equation (involving a recurrence and $\phi(N)$) derived?
Here's the relevant part of the proof, quoting verbatim:


  First, show that $\phi(\phi(n)) &lt; n/2$. This can be done as such:
  
  Let $n = \prod_{i=1}^rp_i^{k_i}$ be the prime factorisation of $n$
  ($p_i$ prime, $k_i&gt;0$)
  
  
  Suppose $n$ is even. Then $\phi(n) = n\prod_{i=1}^r(1-\frac{1}{p_i}) \leq n(1-\frac{1}{2}) \leq n/2.$ Thus $\phi(\phi(n)) &lt; n/2$.
  Suppose $n$ is odd and $n &gt; 1$. Then $\phi(n) = \prod_{i=1}^r (p_i-1)p_i^{k_i-1}$ is even and smaller than $n$. By the previous
  result $\phi(\phi(n)) &lt; n/2$.
  
  
  So we get the desired result.


This lemma is slightly mis-stated, because when $n=2$, $\phi(\phi(n)) = n/2$. But it doesn't invalidate the main result. We will state the lemma as:

Lemma For all $n \ge 2$, $\phi(\phi(n)) \le \frac{n}{2}$.

Tasse's proof sketch is a proof by induction in disguise, but this may not have been obvious to you. I'm going to simplify the proof sketch a little bit and lay it out more like a proper proof by induction.

Proof By induction on $n$.

Base case: If $n=2$, see above.

Inductive step: Suppose that for all $2 \le m &lt; n$, $\phi(\phi(m)) \le \frac{m}{2}$. We want to prove that $\phi(\phi(n)) \le \frac{n}{2}$.

Case 1: Suppose $n$ is even, that is, $n = 2q$ for some $q$. In this case, the multiplicative property of the totient function says that $\phi(2q) = \phi(2) \phi(q) = \phi(q)$.

That is, $\phi(n) = \phi(\frac{n}{2})$. However, another useful property of the totient function is that for all $m \ge 2$, $\phi(m) \le m-1$. Therefore $\phi(\phi(n)) \le \frac{n}{2} - 1$.

This proves what we wanted for this case, but we have also incidentally proven another useful thing: If $n$ is even, then $\phi(n) \le \frac{n}{2}$. This will be important in the next step.

Case 2: Suppose $n$ is odd. That is, $n = p^k q$ for some odd prime $p$, and some positive integer $k$, and some $q$ which doesn't have $p$ as a divisor. Then:

$\phi(n) = \phi(p^k) \phi(q) = \left(p^k - p^{k-1}\right) \phi(q)$.

Now there are two things you need to notice about the right-hand side.

First off, it's less than $n$. Why? Because clearly $p^k - p^{k-1} &lt; p^k$, and it's a standard property of the totient function that $\phi(q) \le q$. (Note that this is also true if $q = 1$.)

Secondly, it's even. Why? Because if $p$ is odd, $p^k - p^{k-1}$ is even.

So by that incidental result we discovered in case 1:

$\phi(\phi(n)) = \phi(\left(p^k - p^{k-1}\right) \phi(q)) \le \frac{\left(p^k - p^{k-1}\right) \phi(q)}{2} &lt; \frac{n}{2}$.

QED