TITLE: Proof - Set of all subsets of a finite set $\Omega$, $2^\Omega$, is also finite
QUESTION [5 upvotes]: As the title implies, I am asked to show that the set of all subsets of some finite set $\Omega$, $2^\Omega$, is also finite. However, I am not really sure if the proof that I've come up with is entirely correct. 
This is what I have:
Since $\Omega$ is finite, then $\exists n \in \mathbb{Z}_+$ such that we can construct a bijection from the finite set of integers $\{x \in \mathbb{Z}_+ | x \leq n\}$ to every element of $\Omega$. 
Furthermore, since $2^\Omega$ is the set of all subsets of $\Omega$, then the total number of such sets is $q = 2^n < \infty$. Therefore, we can construct a bijection from the finite set of integers $\{ x \in \mathbb{Z}_+ | x \leq q \}$ to every element of $2^\Omega$. 
$\square$
I would appreciate any corrections or comments.

REPLY [4 votes]: Furthermore, since $2^\Omega$ is the set of all subsets of $\Omega$, then the total number of such sets is $q=2^n$

Well if you already knew this, you're done!
Remember that $X$ is finite iff there is some bijection $f:[n]\to X$ for some $n\in\mathbb{N}$ where $[n]=\{0,\dotsc,n-1\}$. So that statement implies $2^\Omega$ is finite.
But if you didn't already know $\lvert2^\Omega\rvert=2^{\lvert\Omega\rvert}$ for $\Omega$ finite, then you should prove this by induction on $\Omega$. This is partially the reason we use the notation $2^X$ instead of $P(X)$ for the power set of $X$.