An apparently harmless exercise concerning induction Let $b \in \mathbb{R}, b \ge 2$. Prove by induction that $$(b^n - 1)(b^n - b)(b^n -b^2)\cdots(b^n - b^{n-2}) \ge b^{n(n-1)}-b^{n(n-1)-1}$$ for all $n \in \mathbb{N}, n \ge 1$.

For the case $n = 2$, I have to show that $$b^2 - 1 \ge b^2 - b$$ which is true for $b \ge 1$.
Suppose the result holds for $k \ge 2$, I want to show that it holds for $k + 1$ too.
I know that $$(b^k - 1)(b^k - b)(b^k -b^2)\cdots(b^k - b^{k-2}) \ge b^{k(k-1)}-b^{k(k-1)-1}$$ thanks to the inductive hypothesis, and I want to prove that $$(b^{k+1} - 1)(b^{k+1} - b)(b^{k+1} -b^2)\cdots(b^{k+1} - b^{k-1}) \ge b^{(k+1)k}-b^{(k+1)k-1}.$$
The LHS is $$(b^{k+1} - 1)(b^{k+1} - b)(b^{k+1} -b^2)\cdots(b^{k+1} - b^{k-1}) =$$$$ (b^{k+1} - 1)b(b^k-1)b(b^k-b)b(b^k-b^2)\cdots b(b^k-b^{k-2})=$$$$ b^{k-1}(b^{k+1} - 1)(b^k-1)(b^k-b)(b^k-b^2)\cdots(b^k-b^{k-2}) \ge$$$$ b^{k-1}(b^{k+1} - 1)(b^{k(k-1)}-b^{k(k-1)-1}) =$$$$ b^{k^2+k}-b^{k^2+k-1}-b^{k^2-1}+b^{k^2-2}.$$
The RHS is $$b^{(k+1)k}-b^{(k+1)k-1}= b^{k^2 + k} - b^{k^2 + k - 1}.$$
So I am left to prove that $$b^{k^2+k}-b^{k^2+k-1}-b^{k^2-1}+b^{k^2-2} \ge b^{k^2 + k} - b^{k^2 + k - 1} \Leftrightarrow b^{k^2-2} \ge b^{k^2-1}$$ which is unfortunately not true.
 A: Change variable to $c = \frac1b$. The inequality at hand (the version for $b \ge 2)$ is equivalent to
$$\prod_{k=2}^n (1-c^k) \ge 1 - c\quad\text{ for }\;n \ge 2, c \in \left(0,\frac12\right]$$
To prove this by induction, we will use a stronger form of induction statement.
For $n \ge 1$ and $c \in (0,\frac12]$, let  $\mathcal{S}_n$ be the statement
$$\mathcal{S}_n :\quad  \prod_{k=2}^n (1-c^k) \ge \frac{1-c}{1-c^n}$$
When $n = 1$, LHS is an empty product and evaluates to $1 = $ RHS, so $\mathcal{S}_1$ is true.
Assume $\mathcal{S}_{n}$ is true, we have
$$\prod_{k=2}^{n+1} (1-c^k) = (1-c^{n+1})\prod_{k=2}^n (1-c^k)
\ge (1-c^{n+1})\frac{1-c}{1-c^n}$$
Since $c \in (0,\frac12]$, we have
$$(1-c^{n+1})^2 > 1 - 2c^{n+1} \ge 1 - c^n
\implies \frac{1-c^{n+1}}{1-c^n}  > \frac{1}{1-c^{n+1}}$$
This leads to
$$\prod_{k=2}^{n+1} (1-c^k) > \frac{1-c}{1-c^{n+1}}$$
and hence $\mathcal{S}_n \implies \mathcal{S}_{n+1}$. By induction, $\mathcal{S}_n$ is true for all $n \ge 1$.
As a result, 
$$\prod_{k=2}^n (1-c^k) \ge \frac{1-c}{1-c^n} > 1 - c\quad\text{ for }\; n \ge 1, c \in \left(0,\frac12\right]$$
As a side note, the conjecture about $b \ge 1$ doesn't work in general. 
For example, the inequality for $n = 3$ fails when $b < 
\frac{\sqrt[3]{9+\sqrt{69}} + \sqrt[3]{9-\sqrt{69}}}{\sqrt[3]{18}} \approx 1.324717957244746$.
