How check if the sequence $x_n=(1+\frac{1}{n})^{\frac{1}{n}}$ is monotonically increasing or not? I want to check if the sequence $(1+\frac{1}{n})^{\frac{1}{n}}$ is monotonically increasing. I tried computing $\frac{x_{n+1}}{x_n}$ to check if the ratio is less than 1 or greater than 1, but I am unable to simplify:
$$\frac{x_{n+1}}{x_n} = \frac{n^{\frac{1}{n}}\left(n+2\right)^{\frac{1}{n+1}}}{\left(n+1\right)^{\frac{2n+1}{n\left(n+1\right)}}}$$
I also tried $x_{n+1} - x_n = \left(\frac{n+2}{n+1}\right)^{\frac{1}{n+1}}-\left(\frac{n+1}{n}\right)^{\frac{1}{n}}$
Is there any other way to check the monotonic behaviour of the sequence $x_n$?
Also I would like to know if I can check $y_n=(1-\frac{1}{n})^{\frac{1}{n}}$ using similar arguments?

Here is a simpler solution (which does not use any advanced calculus).
Note that
    \begin{alignat*}{2}
  &&\frac{-1}{n} & < \frac{-1}{n+1}\\
 \Rightarrow &\qquad&1-\frac{1}{n} &< 1-\frac{1}{n+1}\\
 \Rightarrow &\qquad&\left(1-\frac{1}{n}\right)^{\frac{1}{n}} & < \left(1-\frac{1}{n+1}\right)^{\frac{1}{n}}\label{1}\tag{1}
\end{alignat*}
Now,
\begin{alignat*}{2}
  \text{Since,}&\qquad& \left(1-\frac{1}{n+1}\right) & < 1\\
  \Rightarrow &\qquad& \left(1-\frac{1}{n+1}\right)^{\frac{1}{n}} & < 1\\
  \Rightarrow &\qquad&
  \left(1-\frac{1}{n+1}\right)^{\frac{1}{n}}\left(1-\frac{1}{n+1}\right) & <
                                                                       \left(1-\frac{1}{n+1}\right)\\
  \Rightarrow &&\left(1-\frac{1}{n+1}\right)^{\frac{n+1}{n}} & <
                                             \left(1-\frac{1}{n+1}\right)
\end{alignat*}
taking $n+1$-th root,
$$\left(1-\frac{1}{n+1}\right)^{\frac{1}{n}} < \left(1-\frac{1}{n+1}\right)^{\frac{1}{n+1}}\label{2}\tag{2}$$
Thus from (\ref{1}) and (\ref{2}), we get,
    $$a_n=\left(1-\frac{1}{n}\right)^{\frac{1}{n}} <
\left(1-\frac{1}{n+1}\right)^{\frac{1}{n}} <
\left(1-\frac{1}{n+1}\right)^{\frac{1}{n+1}}=a_{n+1}$$
Thus we get $a_{n}<a_{n+1}$. Hence the sequence $(a_n)$ is monotonically increasing.
