Proving that all sets of size n have $2^n$ subsets I've tried to construct a proof for this using recursion. My knowledge of recursion and set theory in general is quite limited so I'd appreciate some feedback!
The claim in symbolic logic: $\forall n \in \mathbb N, \exists u \in U, S(u) \wedge |u| = n \Rightarrow \mathcal |P(u)| = 2^n$ where:
$U:$ the set of everything
$S(x):$ x is a set
Initial Value
Assume $n = 0$
Then $\exists u \in U, |u| = 0$
Then $ u = \varnothing $
Then $ |\mathcal P(u)| = | \{ \varnothing \} | = 1 = 2^0 $
Prove: $\forall n \in \mathbb N, \exists u, x \in U, [S(u) \wedge (x \notin u) \wedge (|u| = n) \wedge \mathcal |P(u)| = 2^n] \Rightarrow  [(|u \cup \{x\}| = n + 1) \wedge (|\mathcal P(u \cup \{x\})| = 2^{n + 1})]$ 
Assume $|u| = n$
Then $ |\mathcal P(u)| = 2^n$
Then $ |u \cup \{x\} | = n + 1$ because  $ |\{x\}| = 1$
Then $ \mathcal |P(u \cup \{x\})| = |\mathcal P(u)| * |\mathcal P(\{x\})| = 2^n * 2 = 2^{n+1} $
Is this convincing enough or do I need to add more? 
EDIT: Fixed notation

you have proved it for sets of size 0
now you have to prove:
if a set of  size n has $ 2^n $ different subsets
then a set of size n+1 has $ 2^{n+1} $ different subsets
and then you can use induction 
