Proof by induction that $(b^n-1)\cdots (b^n-b^{n-2})\ge b^{n(n-1)}-b^{n(n-1)-1}$ Let $b\ge 2$ and $n$ a natural number at least 1.  Prove that
$$(b^n-1)\cdots (b^n-b^{n-2})\ge b^{n(n-1)}-b^{n(n-1)-1}$$

The base-case is obvious.
For the inductive case, assume this holds for $n$ and we show it for $n+1$.  I started by writing $(b^{n+1}-1)\cdots (b^{n+1}-b^{n-1})$ and tried to write a chain of inequalities arriving at $b^{(n+1)n}-b^{(n+1)n-1}$.  This goal is the same as $b^{n(n+1)-1}(b-1)$.
We can factor a $b$ out of every factor other than the first:
$$(b^{n+1}-1)\cdots (b^{n+1}-b^{n-1}) = b^{n-1}(b^{n+1}-1)(b^n-1)(b^n-b)\cdots (b^n-b^{n-2})$$
and apply the inductive hypothesis, so that the above is
$$\ge b^{n-1}(b^{n+1}-1)(b^{n(n-1)}-b^{n(n-1)-1}) $$
$$ = b^{n-1+n(n-1)-1}(b^{n+1}-1)(b-1) $$
$$= b^{n^2-2}(b^{n+1}-1)(b-1) $$
Therefore it suffices to show that
$$ b^{n^2-2} (b^{n+1}-1) \ge b^{n^2+n-1}$$
or
$$b^{n+1}-1\ge b^{n+1}$$
... which ... eh.

Since the only rounding I did was using the inductive hypothesis, it must be that I'm getting a bad result because I'm using the inductive hypothesis in a bad way--like because I'm using it in a product with a large number.  But I don't see an alternate approach.
 A: I'm close,
but it's close to midnight here
so I'll stop with what I have
so far.
I have shown that the insquality
is equivalent to
$\prod_{k=2}^{n}(1-1/b^{k})
\ge 1-1/b
$.
I can't get a proof,
but I'm sure this is true.
Here's my derivation.
Want
$(b^n-1)\cdots (b^n-b^{n-2})\ge b^{n(n-1)}-b^{n(n-1)-1}
$
or
$\prod_{k=0}^{n-2}(b^n-b^k)\ge b^{n(n-1)}-b^{n(n-1)-1}
$
or
$\prod_{k=0}^{n-2}b^k(b^{n-k}-1)
\ge b^{n(n-1)-1}(b-1)
$
or
$b^{\sum_{k=0}^{n-2}k}\prod_{k=0}^{n-2}(b^{n-k}-1)
\ge b^{n(n-1)-1}(b-1)
$
or
$b^{(n-2)(n-1)/2}\prod_{k=0}^{n-2}(b^{n-k}-1)
\ge b^{n(n-1)-1}(b-1)
$
or
$\begin{array}\\
\prod_{k=2}^{n}(b^{k}-1)
&\ge b^{n^2-n-1-(n^2-3n+2)/2}(b-1)\\
&= b^{(n^2+n-4)/2}(b-1)\\
&= b^{n(n+1)/2-2}(b-1)\\
\end{array}
$
For $n=2$
this is
$b^2-1
\ge b(b-1)
$
which is true.
For $n=3$ this is
$(b^2-1)(b^3-1)
\ge b^{4}(b-1)
$
which is true
for $b \ge 2$.
For $n=4$ this is
$(b^2-1)(b^3-1)(b^4-1)
\ge b^{8}(b-1)
$
which is true
for $b \ge 2$.
Since
$\sum_{k=2}^n k
=n(n+1)/2-1
$,
dividing by
$b^{n(n+1)/2-1}
$,
this is
$\prod_{k=2}^{n}(1-1/b^{k})
\ge 1-1/b
$.
Letting $x=1/b$ this is
$\prod_{k=2}^{n}(1-x^k)
\ge 1-x
$
for $0 < x \le 1/2
$
or
$\dfrac1{1-x}
\ge \dfrac1{\prod_{k=2}^{n}(1-x^k)}
$
or
$\dfrac1{(1-x)^2}
\ge \dfrac1{\prod_{k=1}^{n}(1-x^k)}
$.
The right side looks like
the power series
for restricted partitions,
but I don't know where to go from here,
so I'll stop.
