What is the $\lim_{n\rightarrow \infty }(1+\frac{1}{n})^{n^n}$ What is the $$\lim_{n\rightarrow \infty }\left(1+\frac{1}{n}\right)^{n^n}$$
I know that the  $\lim_{n\rightarrow \infty }\left(1+\frac{1}{n}\right)^{n}=e$, so I wanted to find the limit by the same way by taking $\log$ two times  but I found the solution became not easy. Any help
 A: Hint:Observe that: $$\left(1+\dfrac{1}{n}\right)^{n^n}> \left(\left(1+\dfrac{1}{n}\right)^n\right)^n> 2^n$$ for sufficiently large $n$.
A: Here is the solution, maybe, you are looking for: 
Let $L=\lim_{n\rightarrow \infty }\left( 1+\frac{1}{n}\right) ^{n^{n}},$ then
\begin{eqnarray*}
\log L &=&\log \lim_{n\rightarrow \infty }\left( 1+\frac{1}{n}\right)
^{n^{n}} \\
&=&\lim_{n\rightarrow \infty }\log \left( 1+\frac{1}{n}\right) ^{n^{n}} \\
&=&\lim_{n\rightarrow \infty }n^{n}\log \left( 1+\frac{1}{n}\right)  \\
&=&\lim_{n\rightarrow \infty }n^{n-1}\times n\log \left( 1+\frac{1}{n}%
\right)  \\
&=&\lim_{n\rightarrow \infty }n^{n-1}\times \frac{\log \left( 1+\frac{1}{n}%
\right) }{\frac{1}{n}} \\
&=&\lim_{n\rightarrow \infty }n^{n-1}\times \lim_{n\rightarrow \infty }\frac{%
\log \left( 1+\frac{1}{n}\right) }{\frac{1}{n}} \\
&=&\infty ^{\infty }\times 1=\infty .
\end{eqnarray*}
Then
\begin{equation*}
L=e^{\infty }=\infty .
\end{equation*}
Here I have used the following standard limit
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\log \left( 1+x\right) }{x}=1.
\end{equation*}
A: Let $$A_n=\left(1+\frac{1}{n}\right)^{n^n}$$ Taking logarithms $$\log(A_n)=n^n \log\left(1+\frac{1}{n}\right)$$ Now consider Taylor for $\log(1+x)$ when $x$ is small; in the expression, replace $x$ by $\frac 1n$ to get $$\log\left(1+\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3 n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$ $$\log(A_n)=n^{n-1}n\Big(\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3 n^3}+\cdots\Big)=n^{n-1}\Big(1-\frac{1}{2 n}+\frac{1}{3 n^2}+\cdots\Big)$$ So, for large $n$, $$A\approx e^{n^{n-1}}$$
A: We have, $$ \lim_{n\rightarrow \infty }n^{n-1}=\lim_{n\rightarrow \infty }\exp(
\frac{n-1}{n}n\ln n) = \infty~~~\text{and}~~~\lim_{n\rightarrow \infty }\ln\left(1+\frac{1}{n}\right)^{n}  =1 $$
Therefore,
$$\lim_{n\rightarrow \infty }\left(1+\frac{1}{n}\right)^{n^n} =\lim_{n\rightarrow \infty }\exp\left(n^{n-1}\ln \left(1+\frac{1}{n}\right)^{n} \right) =\infty$$
