回忆 $\ln(1+x)$ 的 Taylor 展式 (见 问题1705)
\[
\ln(1+x)=x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+\frac{1}{5}x^5-\cdots,
\]
因此,
\[\ln(1+\frac{1}{x})=\frac{1}{x}-\frac{1}{2}\cdot\frac{1}{x^2}+o(\frac{1}{x^2}).\]
根据题意,
\[
\lim_{x\rightarrow\infty}\Bigl[x\ln(1+\frac{1}{x})\Bigr]^{kx}=e,
\]
于是
\[
\begin{split}
e&=\lim_{x\rightarrow\infty}\Bigl[x\cdot\bigl(\frac{1}{x}-\frac{1}{2}\cdot\frac{1}{x^2}+o(\frac{1}{x^2})\bigr)\Bigr]^{kx}\\
&=\lim_{x\rightarrow\infty}\Bigl[1-\frac{1}{2x}+o(\frac{1}{x})\Bigr]^{kx}\\
&=\lim_{x\rightarrow\infty}e^{kx\ln(1-\frac{1}{2x}+o(\frac{1}{x}))}\\
&=\exp\bigl(\lim_{x\rightarrow\infty}kx\ln(1-\frac{1}{2x}+o(\frac{1}{x}))\bigr)\\
&=\exp\bigl(\lim_{x\rightarrow\infty}kx\cdot(-\frac{1}{2x}+o(\frac{1}{x}))\bigr)\\
&=\exp\bigl(\lim_{x\rightarrow\infty}(-\frac{k}{2}+o(k))\bigr)\\
&=\exp(-\frac{k}{2})=e^{-\frac{k}{2}}
\end{split}
\]
这推出 $k=-2$.