Intuition behind fact that $\lim_{n \to \infty}\frac1 {(1+\frac1 n )^n}=\lim_{n \to \infty}(1-\frac1 n)^n$ Why does
$$\lim \limits_{n \to \infty}\frac1 { \left(1+\frac1 n \right)^n}=\lim \limits_{n \to \infty} \left(1-\frac1 n \right)^n$$
I understand that the math works out because both can be proven to be equal to
$$1-{1\over1!} + {1\over2!} - {1\over3!}... {(-1)^n\over n!}$$ 
I am just wondering if there is any more intuitive explanation.
 A: $$ \left( 1 + \frac{1}{n} \right)^n \left( 1 - \frac{1}{n} \right)^n = \left( 1 - \frac{1}{n^2} \right)^n, $$
which converges to $1$ (take logs, or use the binomial theorem with a remainder term, or Bernoulli's inequalities...)
A: It's a matter of simple algebra. Just note that $$\left(1-\frac {1}{n}\right)^n=\dfrac{1}{\left(\dfrac{n}{n-1}\right)^n}=\frac{n-1}{n}\cdot\dfrac{1}{\left(1+\dfrac{1}{n-1}\right)^{n-1}}$$ Taking limit gives you the desired equality. Using the same technique (and slightly more effort) you can prove that if $x$ is a rational number then $$\lim_{n\to \infty} \left(1+\frac{x}{n}\right)^n=\left(\lim_{n\to \infty} \left(1+\frac{1}{n}\right) ^n\right) ^x$$ (your current question being about $x=-1$).
A: $$\frac{1}{1+\frac{1}{n}} = \frac{n}{n+1} = 1-\frac{1}{n+1}$$
and $\left( 1-\frac{1}{n+1} \right)^n$ and $\left( 1-\frac{1}{n} \right)^n$ behave the same way in the limit. Notice that $\frac{1} { \left(1+\frac1 n \right)^n}=\left( \frac{1} { 1+\frac1 n } \right)^n$ so I can do this.
