I read this paper looking for the best upper bound for computing the Homfly polynomial.
At the end the author makes a substitution which I would like to understand correctly. In Corollary 20 the theorem is
$$O((8k)!^4 \cdot \mathrm{poly}(n)).$$
Then a substitution is done according to $ k = O( \sqrt n ) $ and the final result wasn't exactly what I immediately see is the next step:
$$ e^{ O ( \sqrt n \cdot \log n)}.$$
How is the substitution done? I could assume that $e$ means the base of the natural logarithm but then I would expect $ \ln n $ to appearn instead of $ \log n $ in the exponent. I would think that the step is done by taking the logarithm of the whole expression, simplifying it and then write it with base $e$ but I don't see where $\mathrm{poly}(n)$ went because that could be a very high-order polynom but maybe that becomes a constant after the logarithm is taken. But I still don't understand the steps of the substitution.
One has the crude bound $m! \leq m^m = e^{m \ln m}$, from which $$ (8k)! \leq e^{ 8k\ln(8k)} $$ and $$ (8k)!^4 \leq e^{ 32k\ln(8k)} = e^{O( k\log k)} $$ Plugging in $k = O(\sqrt{n})$, $k\log k = \frac{1}{2}\sqrt{n}\log n$ and so $$ (8k)!^4 = e^{ O(\sqrt{n}\log n)} \,. $$ Multiplying by $\operatorname{poly(n)} = e^{O(\log n)}$ just adds a negligible term in the exponent (even $n^{100}$ is only $e^{100\ln n}$, and $100\ln n = o(\sqrt{n}\log n)$), and so $$ (8k)!^4\operatorname{poly(n)} = e^{ O(\sqrt{n}\log n)} \,. $$
About the base of the logarithm: Note that $\log, \ln, \log_{10}$ all only differ by a constant factor (change of base), and since the $O(\cdot)$ is hiding constant factors, the particular (constant) base of the logarithm is not important when writing $O(\log f(n))$.