\begin{align*} \lim_{x\to\infty}\frac{e^x}{[(1+1/x)^x]^x} &=\lim_{x\to\infty}\frac{e^x}{(1+1/x)^{x^2}}\\ &=\exp\lim_{x\to\infty}\left[x-x^2\ln\left(1+\frac1x\right)\right]\\ &=\exp\lim_{x\to\infty}x^2\left[\frac1x-\ln\left(1+\frac1x\right)\right]\\ &=\exp\lim_{x\to\infty}\frac{1/x-\ln(1+1/x)}{1/x^2}\qquad\left[\sim\frac00\right]\\ &\buildrel\rm L\mathchar"0027H\over=\exp\lim_{x\to\infty}\frac{-1/x^2-[x/(x+1)]\cdot(-1/x^2)}{-2/x^3}\\ &=\exp\lim_{x\to\infty}\left(-\frac{1}{x^2}+\frac{1}{x(x+1)}\right)\left(-\frac{x^3}{2}\right)\\ &=\exp\lim_{x\to\infty}\frac{x}{2x+2}\\ &=\exp\left(\frac12\right)\\ &=\sqrt e. \end{align*}