How to solve for $N$ in $\prod\limits_{k=1}^{N}\left(1-\frac{k}{b}\right) = P$ As the title states, I'm trying to solve for $N$ in

$$\displaystyle \prod_{k=1}^{N}\left(1-\frac{k}{b}\right) = P$$

where $0 &lt; P &lt; 1$ and $b &gt; N$ and is an integer.

I'm trying to solve this for extremely large values of $b$ and $N$ ($\sim 10^{50}$) so a brute-force numerical approach won't work in any reasonable time. I also can't make approximations for very small values of $P$. I'd like $P$ to be able to take on any value between $0$ and $1$.

Can this be solved? If not are there certain approximations that can be made?
If $k$ is fixed, and $b$ is large, then the product is approximately equal to $e^{k N/b},$ so that 
$$\frac{N}{b} = \frac{P}{k},$$
so $$N = \frac{Pb}{k}.$$

Rewrite your LHS as 

$$LHS = \prod_{k=1}^N \frac{b-k}{b} = \frac{(b-1)!}{(b-N-1)! b^N}.$$

Now use Stirling's approximation, to get

$$LHS\approx \frac{\sqrt{2\pi(b-1)} (b-1)^{b-1}/e^{b-1}}{b^N (b-1-N)^{b-N-1}/e^{b-N-1}} = \sqrt{\frac{b-1}{b-N-1}}\frac{(b-1)^{b-1} e^N}{b^N (b-N-1)^{b-N-1}}.$$ Now, you have to decide whether $b$ is a lot or a little bigger than $N.$ The first option is easier.
If $k$ is fixed, and $b$ is large, then the product is approximately equal to $e^{k N/b},$ so that 
$$\frac{N}{b} = \frac{P}{k},$$
so $$N = \frac{Pb}{k}.$$

Rewrite your LHS as 

$$LHS = \prod_{k=1}^N \frac{b-k}{b} = \frac{(b-1)!}{(b-N-1)! b^N}.$$

Now use Stirling's approximation, to get

$$LHS\approx \frac{\sqrt{2\pi(b-1)} (b-1)^{b-1}/e^{b-1}}{b^N (b-1-N)^{b-N-1}/e^{b-N-1}} = \sqrt{\frac{b-1}{b-N-1}}\frac{(b-1)^{b-1} e^N}{b^N (b-N-1)^{b-N-1}}.$$ Now, you have to decide whether $b$ is a lot or a little bigger than $N.$ The first option is easier.