By Using the Binomial expansion as follow that the expression for the first three term is mathematically correct or not? Let the expression is defined as $(1-\frac{1}{i})^{2i}$.
For example: Using the Binomial expansion as follow  $$(1-\frac{1}{i})^{2i}=\sum_{n=0}^{n}\frac{(2i)!}{n!(2i-n)!}(-1)^n(\frac{1}{i})^n$$
$$\approx1-\frac{1}{i}+O((-\frac{1}{i})^3)$$.
Please need your help that the above expression for the first three term is mathematically correct or not, where the big $O$ notation means that the first neglected or unknown term is of the order $(-\frac{1}{i})^3$.
 A: The terms I get are the followings: $$\left(1-\dfrac{1}{i}\right)^{2i}=\sum\limits_{n=0}^{2i}\binom{2i}{n}(-1)^n\dfrac{1}{i^n}=\overbrace{\underbrace{\binom{2i}{0}(-1)^0\dfrac{1}{i^0}}_{1\cdot 1\cdot 1=1}}^{1^{st}}+\overbrace{\underbrace{\binom{2i}{1}(-1)^1\dfrac{1}{i^1}}_{2i\cdot (-1)\cdot 1/i=-2}}^{2^{nd}}+\overbrace{\underbrace{\binom{2i}{2}(-1)^2\dfrac{1}{i^2}}_{i(2i-1)\cdot 1\cdot 1/i^2=\frac{i(2i-1)}{i^2}}}^{3^{rd}}+\ldots$$ The second term does not seem to depend on $i$.
A: That's fairly straight forward.  When n= 0, n!= 0!= 1 and 2i- n= 2i so (2i- n)!= 2i!.  The coefficient is $\frac{(2i)!}{n!(2i- n)!}= \frac{(2i)!}{(2i)!}= 1$.  Of course, $(-1)^0= 1$ and $\left(\frac{1}{i}\right)^01= 1$ so the first term is 1
When n= 1, n!= 1!= 1 and $\frac{(2i)!}{(2i- n)!}$$= \frac{2i(2i-1)!}{(2i-1)!}= 2i$, $(-1)^1= -1$ and $\left(\frac{1}{i}\right)^1= \frac{1}{i}$ so the second term is $-2$, not $-\frac{1}{i}$.
When n= 2, n!= 2!= 2 and $\frac{(2i)!}{(2i-2)!}= $$\frac{2i(2i-1)(2i-2)!}{(2i-2)!}= $$2i(2i-1)$,  $(-1)^2= 1$, and $\left(\frac{1}{i}\right)= \frac{1}{i^2}$ so the third term is $\frac{2(2i)(2i-1)}{i^2}= \frac{4(2i- 1)}{i}$.
