I'm trying to work through this proof and I'm stuck at the fourth equation down on page 2. Firstly, there is a typo in that the coefficient $\sqrt{2\pi}$ should be $1/(\sqrt{2\pi})$. I believe there is a second typo which is that it should read $\ll_\varepsilon e^{|s|^{1+\varepsilon}}$ not $\ll_\varepsilon e^{|s|^\varepsilon}$ (empirical results using limits support this). The first typo is inconsequential, but assuming the second corrected typo I'm now trying to prove $$\left|e^{-(s-1/2)\log(s)}e^s\right|\ll_\varepsilon e^{|s|^{1+\varepsilon}},\tag{*}$$ for all $\varepsilon>0$, where $\text{Re}(s)\geq\delta>0$, presumably for some $\delta>0$.
I manipulated the left hand side, $$|e^{-(s-1/2)\log s+s}|=|e^{-(\sigma-1/2)\log|s|+t\text{Arg}(s)+\sigma}|.$$ So we need to show for some $C>0$ that $$e^{-(\sigma-1/2)\log|s|+t\text{Arg}(s)+\sigma}<Ce^{|s|^{1+\varepsilon}}.$$ I'm stuck at this point and can't seem to prove the inequality. Any help or a hint would be appreciated.
Write $C = e^c$ to get things down from the exponents. You want to show that for a sufficiently large $c$ (depending on $\varepsilon$) the inequality $$-(\sigma - 1/2)\log \lvert s\rvert + t\operatorname{Arg}(s) + \sigma \leqslant \lvert s\rvert^{1 + \varepsilon} + c$$ holds throughout the half-plane $\sigma \geqslant \delta$.
For this we can be pretty brutal. Write $x = \lvert s\rvert$ and note that $\sigma \leqslant x$, $\lvert t\rvert \leqslant x$, and the argument is bounded (in absolute value) by $\pi/2$. Thus it suffices to see that $$(x + 1/2) \lvert\log x\rvert + \pi/2 x + x \leqslant x^{1+\varepsilon} + c$$ holds on the interval $[\delta, +\infty)$ if $c$ is sufficiently large. But $$\lim_{x \to +\infty} \frac{\log x}{x^{\varepsilon}} = 0$$ implies $$\lim_{x \to +\infty} \:\bigl((x+1/2)\lvert \log x\rvert + (1 + \pi/2)x - x^{1+\varepsilon}\bigr) = -\infty\,,$$ so the function on the left is bounded from above, as we wanted to show.
To see that it's indeed a typo and $e^{\lvert s\rvert^{\varepsilon}}$ on the right hand side does not work, consider $s = \frac{1}{2} + it$. For large $\lvert t\rvert$ we have $$\Bigl\lvert e^{-(s-1/2)\log s + s}\Bigr\rvert \geqslant e^{\lvert t\rvert}$$ but $$e^{\lvert s\rvert^{\varepsilon}} < e^{(2\lvert t\rvert)^{\varepsilon}}\,,$$ which shows the inequality $$\Bigl\lvert e^{-(s-1/2)\log s + s}\Bigr\rvert \leqslant Ce^{\lvert s\rvert^{\varepsilon}}$$ cannot hold for $\varepsilon < 1$.