Showing convergence to $\frac{1}{e}$ I am trying to show that the sequence $$a_n =  (1-\frac{1}{n})^n$$ converges to $\frac{1}{e}$.
So far, I have started with the fact that $$\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n = e$$
Through some manipulation, I've gotten to showing that by taking reciprocals on both sides, we can say that $$\lim_{n\rightarrow\infty}(1-\frac{1}{n+1})^n = \frac{1}{e}$$
Did I make an algebraic error? Or is there a way to definitively say that having $n+1$ in my second numerator does not matter?
 A: We begin with the limit definition of $e$ written as 

$$\bbox[5px,border:2px solid #C0A000]{e\equiv \lim_{n\to \infty}\left(1+\frac1n\right)^n} \tag 1$$


Next, we note that 

$$\left(1+\frac1n\right)^n\left(1-\frac1n\right)^n=\left(1-\frac1{n^2}\right)^n \tag 2$$

It is evident that the right-hand side of $(2)$ is bounded above by $1$.  Then, using Bernoulli's Inequality, we find that 

$$\left(1-\frac1{n}\right) \le \left(1-\frac1{n^2}\right)^n \le 1$$

whereupon application of the squeeze theorem yields

$$\lim_{n\to\infty}\left(1-\frac1{n^2}\right)^n =1 \tag 3$$


Putting together $(1)-(3)$, we find that 

$$\begin{align}
1&=\lim_{n\to\infty}\left(1-\frac1{n^2}\right)^n\\\\
&\lim_{n\to \infty}\left(1+\frac1n\right)^n\left(1-\frac1n\right)^n\\\\
&=\left(\lim_{n\to \infty}\left(1+\frac1n\right)^n\right)\,\left(\lim_{n\to \infty}\left(1-\frac1n\right)^n\right)\\\\
&=e\,\lim_{n\to \infty}\left(1-\frac1n\right)^n\\\\
e^{-1}&=\lim_{n\to \infty}\left(1-\frac1n\right)^n
\end{align}$$

as was to be shown!
A: Note that
$$
\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)^{n+1}=\lim_{n\to\infty}\left[\left(1-\frac{1}{n+1}\right)\left(1-\frac{1}{n+1}\right)^n\right]=1\cdot\frac{1}{e}=\frac{1}{e}.
$$
Now, let $m=n+1$, and note that $m\to\infty$ as $n\to\infty$.
A: Let it be 
$$
a_n=(1+1/n)^n=:1+a_n',\qquad b_n=(1-1/n)^n 
$$
Since you already know that $a_n\to e$, it is sufficient to prove that $a_n b_n\to 1$.
Now
$$
a_n b_n=\left(1-\frac{1}{n^2}\right)^n=1-c_n,
$$
where
$$
|c_n|\leq\sum_{k=1}^{n}\left(\frac{1}{n^2}\right)^k\binom{n}{k}\leq\frac{1}{n}\cdot \sum_{k=1}^{n}\left(\frac{1}{n}\right)^k\binom{n}{k}=\frac{a_n'}{n} 
$$
Since $a_n'\to e-1$, it follows $c_n\to 0$.
