The expansion of $(1+\frac{1}{n})^n$, explanation needed.

Question

Here is the binomial expansion: $$(1+\frac{1}{n})^n=\frac{n\cdot\frac{1}{n}}{1!}+\frac{n(n-1)(\frac{1}{n})^2}{2!} + \frac{n(n-1)(n-2)(\frac{1}{n})^3}{3!}+ ... + \frac{n(n-1)(n-2)...(n-(n-1)(\frac{1}{n})^n}{n!}=1+1+\frac{1}{2!}(n^2-n)(\frac{1}{n})^2+\frac{1}{3!}\frac{(n^2-n)}{n^2}\frac{(n-2)}{n}+...+\frac{1}{n!}\frac{(n^2-n)}{n^2}\frac{(n-2)}{n}$$ I don't understand why $(\frac{1}{n})^3$ and $(\frac{1}{n})^n$ has disappeared and instead theeach multiplier is divided by $n^2$ and $n$. Can someone explain?

Answer 1

Actually, what you wrote has so many mistakes that it is better to start all over again.

You have\begin{align}\left(1+\frac1n\right)^n&=\binom n0+\binom n1\frac1n+\binom n2\left(\frac1n\right)^2+\binom n3\left(\frac1n\right)^3+\cdots+\binom nn\left(\frac1n\right)^n\\&=1+n\frac1n+\frac{n(n-1)}{2!}\cdot\frac1{n^2}+\frac{n(n-1)(n-2)}{3!}\cdot\frac1{n^3}+\cdots+\frac{n!}{n!}\cdot\frac1{n^n}\\&=1+1+\frac1{2!}\frac{n-1}n+\frac1{3!}\frac{n-1}n\frac{n-2}n +\cdots+\frac1{n!}\frac{n-1}n\frac{n-2}n\cdots\frac1n.\end{align}