Showing $\lim_{m\rightarrow\infty} \left(1-\frac1{m^2}\right)^m=1$ I want to prove $$\lim_{m\rightarrow\infty} \left(1-\frac1{m^2}\right)^m=1$$ without using the fact that $\lim_{m\rightarrow\infty}\left(1+\frac1m\right)^m=\mathrm e$.
I know by the Bernoulli-Inequality $$\left(1-\frac1{m^2}\right)^m\geq1-\frac1m$$
But now I don't know how to show $\left(1-\frac1{m^2}\right)^m\leq1$ for all $m\in\mathbb N$.
Anybody could help? Thanks.
 A: When you get stuck, try proving something harder instead. Perhaps
$$\left( 1-\frac1{a}\right)^n \le 1 $$
for all $a\ge 1$ and $n\ge 1$?
A: $$\left(1-\frac{1}{m^2}\right)^m=\left(1-\frac{1}{m}\right)^m\times \,\,\left(1+\frac{1}{m}\right)^m$$
A: Apply $\ln$, rewrite with $1/m$ in the denominator and then use L'Hospital's rule to bring the expression into the form of a rational function.
A: I have proof on this site: $e^x=\displaystyle\lim_{n \to \infty}(1+\frac{x}{n})^n$
-here I am gonna modificate some elements and others not explaning .
So, we gonna need this binomial expansion:
$(1-y)^n=\binom{n}{0}y^0-\binom{n}{1}y^1+\binom{n}{2}y^2+...+(-1)^{n-1}\binom{n}{n-1}y^{n-1}+(-1)^{n}\binom{n}{n}y^n$
Hense:
$(1-y)^n=1-n*y+\frac{(n-1)n}{2!}y^2+...+(-1)^{n-1}*n*y^{n-1}+(-1)^{n}*y^n$
Here: $n=m$ and $y=\frac{1}{m^2}$ :
$(1-\frac{1}{m^2})^m=1-m*\frac{1}{m^2}+\frac{(m-1)m}{2!}(\frac{1}{m^2})^2+...+(-1)^{m-1}*m*(\frac{1}{m^2})^{m-1}+(-1)^{m}*(\frac{1}{m^2})^m$
$=1-\frac{1}{m}+\frac{1}{2}\frac{m-1}{m^3}+...=1-\frac{1}{m}+\frac{1}{2}(\frac{1}{m^2}-\frac{1}{m^3})+...$
Is not hard to see here that we have 1 + something (very) small:
$\displaystyle\lim_{n \to \infty}(1-\frac{1}{m^2})^m=1+0+0+...=1$ Q.E.D.
PS: Alternatively we could use:
$(1+y)^n=\binom{n}{0}y^0+\binom{n}{1}y^1+\binom{n}{2}y^2+...+\binom{n}{n-1}y^{n-1}+\binom{n}{n}y^n$
; where  $y=-\frac{1}{m^2}$
A: Since taking $\lim_{m \to \infty}\left( 1 - \frac{1}{m^{2}} \right)^{m}$ gives us the indeterminate form $1^{\infty}$, and, as the problem does not forbid us from using L'H Rule, let's use it here.
First, observe that $$ \ln \left( 1 - \frac{1}{m^{2}} \right)^{m} = m \ln \left( 1 - \frac{1}{m^{2}} \right) = \frac{\ln \left( 1 - \frac{1}{m^{2}} \right)}{\frac{1}{m}} $$
Hence, since the limit as $m \to \infty$ of the RHS of the above equality equals $0$, 
$$ \lim_{m \to \infty}\left( 1 - \frac{1}{m^{2}} \right)^{m} = e^{\lim_{m \to \infty}\left( 1 - \frac{1}{m^{2}} \right)^{m}} = e^{0} = 1. $$
