Let $ A $ be a finite set, and prove that $ \mathcal{P}\left(A\right) $ is also finite.
Here's what I've done:
Since $ A $ is finite, we can assume that $ |A|=n $ for some natural number $ n\in \mathbb{N} $.
From the assumption above, it follows that there exists a bijection $ f:\mathbb{N}^{<n}\to A $.
We'll define $ g:\mathcal{P}\left(A\right)\to\left\{ 0,1\right\} ^{\mathbb{N}^{<n}} $ by:
For any $ B\in\mathcal{P}\left(A\right) $
$ g\left(B\right)\left(m\right)=\begin{cases} 0 & f\left(m\right)\notin B\\ 1 & f\left(m\right)\in B \end{cases} $
I'm sure we all agree that $ g $ is a bijection. And therefore $ |\mathcal{P}\left(A\right)|=|\{0,1\}^{\mathbb{N}^{<n}}| $.
Now, by definition, for any sets $ A,B $ such that $ |A|=\alpha,|B|=\beta $, the cardinality of $ A^B $ defined as $ |A|^{|B|}=\alpha^{\beta} $. In our case, by the definition, $ |\{0,1\}^{\mathbb{N}^{<n}}|=2^{n} $, because $ |\{0,1\}|=2,|\mathbb{N}^{<n}|=n $.
Thus, we get that $ |\mathcal{P}\left(A\right)|=2^{n}\in\mathbb{N} $. And since $ 2^{n}<\aleph_{0} $, we get that $ \mathcal {P}(A) $ is finite.
This proof is legit? Or maybe I've abused the definitions? I'm asking because this question appeared in my exam (it wasn't written that we have to prove by definition of finite set, so I proved my way).
Thanks in advance.
First some remarks: Just fyi, $\mathbb{N}^{<n}$ is frequently used as notation for the set of sequences of natural numbers of length less than $n$ (aka the disjoint union of $\mathbb{N}^{k}$ for $0\leq k<n$) not the set $\{0,1,\ldots,n-1\}$. So readers could be momentarily confused. Also you say "I'm sure we'll all agree that $g$ is a bjection." But this is really the content of the whole proof. The rest about computing the cardinality of $\{0,1\}^{\mathbb{N}^{<n}}$ is just window dressing. If I were grading an exam in which a student only wrote that without actually proving it's a bijection, then I would certainly not give full credit.
Now some more specific analysis of your proof. First, it's completely correct (up to showing all the details as I say above). But you can simplify it. For example, there is no reason to invoke the bijection $f$. $A$ is a finite set and so you could just as well define $g$ from $P(A)$ to $\{0,1\}^{A}$ such that $g(B)(x)=1$ iff $x\in B$. Then $\{0,1\}^{A}$ is finite since $\{0,1\}$ and $A$ are finite (as you say already).
Another approach is to use induction to show that if $A$ has size $n$ then $P(A)$ has size $2^n$. That proof has its benefits, but I like your approach for the combinatorial flavour. Your proof uniquely identifies a subset of $A$ with $|A|$ many yes/no questions (for each element $x$ in $A$, is it in the subset?). So $|A|$ questions, each with $2$ answers, hence $2^{|A|}$ many different ways to answer them all.