Proving that a limit goes to infinity Looking for some help in proving why the following infinity goes to $\infty$ as n approaches $\infty$
$$\frac{(1-\frac{1}{n})^n}{1-(1-\frac{1}{n})}$$
 A: HINT:
So, we have $\lim_{n\to\infty}n\cdot\left[\lim_{n\to\infty}\left(1-\dfrac1n\right)^{-n}\right]^{-1}$
Do you know $\lim_{m\to\infty}\left(1+\dfrac1m\right)^m=e$
A: $$\dfrac{(1-\dfrac{1}{n})^n}{1-(1-\dfrac{1}{n})}=\\
\dfrac{\binom{n}{0}1^{n}{(-\dfrac{1}{n})}^{0}+\binom{n}{1}1^{n-1}{(-\dfrac{1}{n})}^{1}+\binom{n}{2}1^{n}{(-\dfrac{1}{n-2})}^{2}+...+\binom{n}{n}1^{0}{(-\dfrac{1}{n})}^{n}
}{1-(1-\dfrac{1}{n})}=\\
\dfrac{1+n{(-\dfrac{1}{n})}+\dfrac{n(n-1)}{2}{(-\dfrac{1}{n})}^{2}+\dfrac{n(n-1)(n-2)}{6}{(-\dfrac{1}{n})}^{3}+...+1{(-\dfrac{1}{n})}^{n}
}{1-(1-\dfrac{1}{n})}=\\
\dfrac{1-1+\dfrac{(n-1)}{2}{(\dfrac{1}{n})}+\dfrac{(n-1)(n-2)}{6}{(-\dfrac{1}{n^2})}+...+1{(-\dfrac{1}{n})}^{n}
}{\dfrac{1}{n}}=\\$$
so apply limit 
$$\lim_{n \to \infty }\dfrac{1-1+\dfrac{(n-1)}{2}{(\dfrac{1}{n})}+\dfrac{(n-1)(n-2)}{6}{(-\dfrac{1}{n^2})}+...+1{(-\dfrac{1}{n})}^{n}
}{\dfrac{1}{n}}=\\
\lim_{n \to \infty }\dfrac{\dfrac{(n-1)}{2n}-\dfrac{(n-1)(n-2)}{6n^2}+...+1{(-\dfrac{1}{n})}^{n}
}{\dfrac{1}{n}}=\\
\lim_{n \to \infty }n({\dfrac12-\dfrac16+\dfrac{1}{24}-\dfrac{1}{120})
}=\to \infty\\
$$
