Does it hold that $\prod\limits_{i=1}^n(1-p^{-i}) \geq {(1-p^{-1}})^2$?

Question

I was trying to find a lower bound for the product $$\prod\limits_{i=1}^n(1-p^{-i})$$ and it seems that the following inequality holds but I was not able to prove it: for every natural $n$, $p$,

$$\prod_{i=1}^n(1-p^{-i}) \geq {(1-p^{-1}})^2$$

Answer 1

Consider some real number $x$ in $[0,\frac12]$. For every $n\geqslant1$, $$\color{red}{\frac{x^n}{1-x^n}}\leqslant1$$ Using the expansion $$\log(1-t)=-\sum_{n=1}^\infty\frac{t^n}n$$ one sees that $$\sum_{i=2}^\infty\log(1-x^i)=-\sum_{n=1}^\infty\sum_{i=2}^\infty\frac{x^{in}}n=-\sum_{n=1}^\infty\frac{x^n}n\color{red}{\frac{x^n}{1-x^n}}\geqslant-\sum_{n=1}^\infty\frac{x^n}n=\log(1-x)$$ Exponentiating both sides and multiplying everything by $(1-x)$, one gets $$\prod_{i=1}^\infty(1-x^i)\geqslant(1-x)^2$$ which is equivalent to (a strengthening of) the desired inequality for every $p\geqslant2$.