Evaluating $\lim_{n\rightarrow\infty} (\frac{(1+\frac{1}{n})^n}{e})^n$

This is a companion to Eevee Trainer's answer: The same (il)logic that says

$$\left((1+1/n)^n\over e\right)^n\to\left(e\over e\right)^n=1^n=1$$

would also say

$$\left(1+{1\over n}\right)^n\to(1+0)^n=1^n=1$$

Since everyone else has decided to cover ways to calculate the limit (despite a comment from you mentioning that you want to know why you're wrong and not how to solve the exercise), I'll answer focusing on that. First, simplifying, you have

$$\lim_{n \to \infty} \frac{ ((1+1/n)^n)^n }{e^n}$$

In replacing the top expression with $e^n$, you implicitly assume that you can take the limit inside as so, with your substitution in blue:

$$\lim_{n \to \infty} \left( \left( 1 + \frac 1 n \right)^n \right)^n = \left(\color{blue}{\lim_{n \to \infty} \left( 1 + \frac 1 n \right)^n} \right)^n =\color{blue}{e}^n$$

However, you have a dependence on $n$ on the outer parentheses, and thus this step is not justified. You can only move a limit inside a (continuous) function when you're not suddenly moving a dependence on $n$ to the outside.

Let $x=\frac1n$. Using $$ \ln(1+x)=x-\frac12x^2+O(x^3)$$ one has \begin{eqnarray} &&\lim_{n\rightarrow\infty} \ln\bigg(\frac{(1+\frac{1}{n})^n}{e}\bigg)^n\\ &=&\lim_{n\rightarrow\infty} n\bigg(n\ln(1+\frac{1}{n})-1\bigg)\\ &=&\lim_{n\rightarrow\infty} \frac{n\ln(1+\frac{1}{n})-1}{\frac1n}\\ &=&\lim_{x\rightarrow0} \frac{\frac1x\ln(1+x)-1}{x}\\ &=&\lim_{x\rightarrow0} \frac{-\frac12x+O(x^2)}{x}\\ &=&-\frac12. \end{eqnarray} So $$ \lim_{n\rightarrow\infty} \bigg(\frac{(1+\frac{1}{n})^n}{e}\bigg)^n=e^{-1/2}. $$