Proving convergence of real sequence Having already shown that $$\lim_{n\to \infty } \left ( \left (1+\frac{1}{n} \right )^n \right ) $$ exists I am then asked to show that $$\lim_{n\to \infty } \left ( \left (1-\frac{1}{n} \right )^n \right ) $$ exists.
Is there a way of linking the two limits together and the convergence of the first convergence implies the convergence of the second sequence.?
 A: It is enough to prove $\displaystyle\Bigl(1+\frac1n\Bigr)^n\Bigl(1-\frac1n\Bigr)^n$ has a limit.
$$\Bigl(1+\frac1n\Bigr)^n\Bigl(1-\frac1n\Bigr)^n=\Bigl(1-\frac1{n^2}\Bigr)^n<1$$
and by Bernoulli's inequality, we have
$$\Bigl(1-\frac1{n^2}\Bigr)^n\ge 1-n\cdot\frac1{n^2}=1-\frac1n.$$
The squeeze theorem ensures that
$$\lim_n\Bigl(1+\frac1n\Bigr)^n\Bigl(1-\frac1n\Bigr)^n=1,$$
so that $$\lim_n\Bigl(1-\frac1n\Bigr)^n=\frac1{\lim_n\Bigl(1+\dfrac1n\Bigr)^n}.$$
A: $$\lim_{n \to +\infty} \left(1-\frac{1}{n}\right)^n = \lim_{n \to +\infty} \left(\frac{n-1}{n}\right)^n  =  \lim_{n \to +\infty}\frac{1}{\left(\frac{n}{n-1}\right)^n} = \lim_{n \to +\infty}\frac{1}{\left(\frac{n-1+1}{n-1}\right)^n} = \lim_{n \to +\infty}\frac{1}{\left(1+\frac{1}{n-1}\right)^n} = \left(\lim_{n \to +\infty}\frac{1}{\left(1+\frac{1}{n-1}\right)^{n-1}}\right)\left(\lim_{n \to +\infty}\frac{1}{1+\frac{1}{n-1}}\right) = \frac{1}{\lim_{n \to +\infty}\left(1+\frac{1}{n-1}\right)^{n-1}} = \frac{1}{\lim_{m \to +\infty}\left(1+\frac{1}{m}\right)^{m}}$$
where I substituted $m = n-1$.
A: Hint: Write
$$\left(1-\dfrac{1}{n}\right)^n = \dfrac{1}{\left(\dfrac{n}{n-1}\right)^n}$$
