Within the paper PRIMES is in P the following inequality can be found (on page 4, in the proof of Lemma 4.3) $$ n^{\lfloor \log(B) \rfloor} \prod_{i=1}^{\lfloor \log^2(n) \rfloor} (n^i - 1) \hspace{2mm} < \hspace{2mm} n^{\lfloor \log(B) \rfloor + \frac{1}{2} \log^2(n) (\log^2(n) - 1)} $$
This implies that $$ \prod_{i=1}^{\lfloor \log^2(n) \rfloor} (n^i - 1) \hspace{2mm} < \hspace{2mm} n^{\frac{1}{2} \log^2(n) (\log^2(n) - 1)} $$ However, I would have expected this inequality to be
$$ \prod_{i=1}^{\lfloor \log^2(n) \rfloor} (n^i - 1) \hspace{2mm} < \hspace{2mm} n^{\frac{1}{2} \log^2(n) (\log^2(n) + 1)} $$ (plus 1 at the end, rather than minus 1) due to the summation rule $$ \sum_{i=1}^n i = \frac{1}{2} n(n+1) \hspace{5mm} \Rightarrow \hspace{5mm} \prod_{i=1}^n n^i = n^{\frac{1}{2} n(n+1)} $$
Assuming I am wrong about this, how was the inequality formed?
