Let $X_i \quad$ be independent identically distributed Random Variables s.t. $E|X|^q < \infty$ ( some $q \in \mathbb{N} )$. When defining 

$Y_i:= (X_i-\bar{X}_n)^q$

why is it true that $Y_i$ are iid for $i \in (1,..,n)$ ? Does subtracting the sample mean not introduce interdependence ?

There is actually a straightforward reason.   As $99$ is $1$ less than $100$, we get a fairly simple expression for its decimal $$\frac{1}{99}=0.01010101010101\overline{01}\dots$$  Now, $$\frac{1}{9801}=\left(\frac{1}{99}\right)^2,$$ and the decimal expansion follows from the formula for general power series $$\left(\sum_{n=1}^\infty x^n\right)^2= x\sum_{n=1}^\infty nx^n.$$ 

Letting $x=\frac{1}{100}$, the decimal expansion for $\frac{1}{99}$ given above is exactly the same thing as writing $\frac{1}{99}=\sum_{n=1}^\infty x^n$.  Applying our identity, the $x$ in front accounts for the double zero.  Once $n$ is around $99$ we expect to miss a number because we are forcing things to be in decimal, and there will be carrying, which is why the number 98 is missed.

A similar pattern will occur for $\frac{1}{998001}=\left(\frac{1}{999}\right)^2,$ since as before $$\frac{1}{999}=0.001001\overline{001}.$$
