This is part of a proof that I saw in lectures but the lecturer quickly skipped any calculation and just stated this result. It's not obvious to me that this must be true from the formula, although plotting in desmos does convince me. I believe this is true for all $n$.
I think to show this, we need to use $1-p \leq e^{-p}$, and I've had a look at the Pentagonal number theorem, but that seems less useful in this case.
Do you have any idea how to prove this result?
Actually, it holds for $0<p\leq 1$. First, by using the inequality $1-p\leq e^{-p}$ we get that $$p\prod_{k=1}^n (1-(1-p)^k) \geq p\prod_{k=1}^n (1-e^{-kp})$$
Taking logarithms and moving $p$ to the RHS it is sufficient to show that $$\sum_{k=1}^n \log(1-e^{-kp}) \geq -\frac{3}{p}-\log(p)$$
The function $x\mapsto \log(1-e^{-xp})$ is negative and monotonically decreasing in absolute value (for $x > 0$ and fixed $p$) and so from the integral test we get $$\sum_{k=1}^n \log(1-e^{-kp}) \geq \int_0^\infty \log(1-e^{-xp})\,dx=-\frac{\pi^2}{6p}$$
It remains to show that $-\frac{\pi^2}{6p}\geq -\frac{3}{p}-\log(p)$ which is easily verified to be true for $0<p\leq 1$.