Proof of Expected Value for Independent random variables with two possibilities Let ${B_i}$ ($i=$natural number) be a random variable which takes $1$ with probability $p$ and takes $-1$ with probability $1-p$ where $0<p<1$. Assume that ${B_i}$ and ${B_j}$ are independent if $i \neq j$. Let  ${S_N}=\sum_{i=1}^{N} {B_i}$ ($N=$Natural Number) and $E[.]$ denote expectation.
Show that 
$E\left[{B_i}^m{B_j}^n\right]=E\left[{B_i}^m\right]E\left[{B_j}^n\right]$ for any natural numbers $m$ and $n$ if $i\neq j$.
I am really new to this subject, so I hope someone experienced can help show me how I should prove this.

If two random variables $X$ and $Y$ are independent then $E(XY)=E(X)E(Y)$. This result does not depend on the type of random variables.
This is because $$E(XY)=\sum_{x,y}xyP(X=x,Y=y)=\sum_{x,y}xyP(X=x)P(Y=y)\\=\Big(\sum_xxP(X=x)\Big)\Big(\sum_yyP(Y=y)\Big),$$
which uses independence to go between the second and third expressions.
For sums, much more is true: $E(X+Y)=E(X)+E(Y)$ even if the variables are not independent.
The $n$ and $m$ in your question are a distraction. $X_i=\pm1$ so if $m$ is odd then $X_i^m\equiv X_i$, and if $m$ is even then $X^m\equiv E(X^m)=1$.
