At first glance, the formula looks lightly wrong: the last term in the should only be $(1-2^{n-2}/N(s))$, giving
$$
P_{ind}=\prod_{k=1}^{n-1}\left(1-\frac{2^{k-1}}{N(s)}\right)
$$
overall. Thus every term is a half, or larger.
My reasoning is as follows: you perform a measurement and get a random outcome. The first time you do this, it can be any outcome except the all 0 string. This happens with probability $1-1/N(s)$.
The second time, you want any string except the all zeros, or the answer you got last time, $y_1$. Thus, the term $1-2/N(s)$.
The third time, you want any string except the all zeros, $y_1$, $y_2$ or $y_1\oplus y_2$. Thus, the term $1-4/N(s)$.
Once you have $k-1$ linearly independent strings $y_1$ to $y_{k-1}$ and you're trying top find the $k^{th}$, there are $2^{k-1}$ answers you don't want to get: the $2^{k-1}$ answers that are linearly dependent on the strings you already have (note that this counting includes the all zeros string). You keep going until the last term, $k=n-1$ because you're trying to find $n-1$ linearly independent cases.
Incidentally, this is not the way that I would ever make the argument. Who cares about the probability of needing exactly $n-1$ calls? You can just keep repeating as many times as you need to in order to find $n-1$ linearly independent strings. Since we've already argued that the worst-case probability of finding a new linearly independent string is 1/2, this means that, on average, no more than $2(n-1)$ trials would be required (and actually somewhat less, because early on you're far more likely to get a hit). You could also apply a Chernoff bound to prove that the probability of needing significantly more runs than that is vanishingly small. OK, that's essentially where the solution gets to, it just feels a little excessive (to me).
