Derivation in "Step-by-Step... HHL" is unclear

Write \begin{align} \frac{1}{2^n} \sum_{y=0}^{2^n-1} \sum_{k=0}^{2^n-1} e^{2 \pi i k(\phi - y/N)} |y\rangle = \sum_{y=0}^{2^n-1} \left(\frac{1}{2^n} \sum_{k=0}^{2^n-1} e^{2 \pi i k(\phi - y/N)}\right) |y\rangle. \end{align} The stuff in the parentheses is just a coefficient of $|y\rangle$.

Suppose for some $y_0$, we have $\phi - y_0/N = 0$, then the coefficient of $|y_0\rangle$ is $$\left(\frac{1}{2^n} \sum_{k=0}^{2^n-1} e^{2 \pi i k(\phi - y_0/N)}\right) = \left(\frac{1}{2^n} \sum_{k=0}^{2^n-1} e^{2 \pi i k0}\right) = \frac{2^n}{2^n} = 1.$$ Since $\phi - y_0/N =0$, we can solve for $y_0$, \begin{align} \phi - y_0/N &=0\\ \phi &= y_0/N\\ \phi N &= y_0. \end{align} So we have $|y_0 \rangle = |\phi N\rangle$.

Addendum:
Even though this was not part of the question, I added it for completeness.

If $\phi - y/N \neq 0$, then $$\sum_{k=0}^{2^n-1} e^{2 \pi i k(\phi - y/N)} =0.$$ This follows from the geometric series formula: $$1 + \alpha + \alpha^2 + \ldots + \alpha^{n-1} = \begin{cases} \frac{1-\alpha^n}{1-\alpha}, \text{ if } \alpha \neq 1\\ n, \text{ if } \alpha =1 \end{cases}$$ Therefore, if $\phi - y/N \neq 0$, then $\alpha:=e^{2\pi i (\phi - y/N)} \neq 1$. It follows that $$\sum_{k=0}^{2^n-1} e^{2 \pi i k(\phi - y/N)} = \frac{1-e^{2\pi i (\phi - y/N)2^n}}{1-e^{2\pi i (\phi - y/N)}} = 0.$$ This is because we can write $$1-e^{2\pi i (\phi - y/N)2^n} = 1 - (e^{2\pi i 2^n})^{\phi - y/N} = 1 - (1)^{\phi - y/N} = 0.$$