Why is the square of a Bernoulli random variable still a Bernoulli random variable? Suppose that:
$$
X \sim Bern(p)
$$
Then, intuitively $X^2 = X \sim Bern(p)$. However, when I try to think of it logically, it doesn't make any sense. 
As an example, $X$ is $1$ with probability $p$ and $0$ with probability $1-p$. Then, $X^2 = X\cdot X$ is $1$ only when both $X$'s are $1$, which occur with probability $p^2$, and so it doesn't seem like $X^2 = X$. Can someone tell me what is wrong here?
 A: 
As an example, $X$ is $1$ with probability $p$ and $0$ with probability $1-p$. Then, $X^2 = X\cdot X$ is $1$ only when both $X$'s are $1$, which occur with probability $p^2$, and so it doesn't seem like $X^2 = X$. Can someone tell me what is wrong here?

There are not "both $X$'s".   $X^2$ is the product of $X$ and itself.  
Whenever $X$ equals $1$, then $X^2$ must equal $1^2$.   So $\{X=1\}$ and $\{X^2=1^2\}$ are the exact same event, and thus cannot have anything except the same probability measure. $$\mathsf P(X^2=1^2)~=~\mathsf P(X=1)~=~p$$
Likewise, $\{X=0\} = \{X^2=0^2\}$.$$\mathsf P(X^2=0^2)~=~\mathsf P(X=0)~=~1-p$$
A: 
$X^2=X.X$ is $1$ only when both $X$'s are $1$, which occurs with probability $\color{blue}p$. 

Notice that $X$ and $X$ are identical, and dependent. The two $X$'s refer to the same thing. 
Since $X$ takes binary value, we indeed have $X^2=X$.
A: Your logic is flawed. When $X$ is Bernoulli, the events $\{\omega: X^2(w) = 1\}$ and $\{w:X(w) = 1\}$ are precisely the same events. 
