Counting the number of bijections from $I_n$ to $A$ Is the following Proof Correct?
PRELIMANARY INFORMATION


*

*$I_n = \{i\in\mathbf{Z^+}|i\leq n\}$

*$A\thicksim B\Leftrightarrow$ There is a bijection from A to B

*A is finite $\Leftrightarrow\exists n\in\mathbf{N}(A\thicksim I_n)$ 

*(7) Suppose $A$ and $B$ are sets and $A$ is finite. Prove that $A ∼ B$ iff $B$ is also finite and $|A| = |B|$.

*(19) Given that $n$ is a positive integer and for each $i\in I_n$, $A_i$ is a finite set and $\forall i\in I_n\forall j\in I_n(i\neq j \implies A_i\cap A_j = \varnothing)$. $A_i$ is finite and 
$$|\cup_{i\in I_n}A_i| = \sum_{i=1}^{n}|A_i|$$
Theorem. Let $|A| = n$ and $F = \{f|f:I_n \xrightarrow{\rm bijection} A\}$ then $F$ is finite and $|F| = n!$
Proof. We construct the proof by recourse to Mathematical-Induction on $|A| = n$.
For $|A| = 0$ it is evident that $A = I_0 = \varnothing$ consequently the $|F| = 1 = 0!$
Now Assume for an arbitrary natural number $k$ that given any set $X$ such that $|X| = k$ the set $Y = \{f|f:I_n \xrightarrow{\rm bijection}X\}$ is such that $|Y| = k!$
Let $A$ be an arbitrary set and assume that $|A| = k+1$, furthermore let $F = \{f|f:I_n \xrightarrow{\rm bijection} A\}$ and $h:I_{k+1}\xrightarrow{\rm bijection}A$. Now consider the following definition.
$$\forall i\in I_{k+1}\left(A_{i} = \{g|g:I_k\xrightarrow{\rm bijection}A\backslash\{h(i)\}\}\right)$$
We now prove a series of Lemmas that will help with our subsequent argument.
Lemma.(1). $\forall i\in I_{k+1}\forall j\in I_{k+1}(i\neq j\implies A_i\cap A_j = \varnothing)$
Proof. Assume on the contrary that for some $i,j\in I_{k+1}$ that $i\neq j$ but $A_i\cap A_j \neq\varnothing$ consequently there exists a $s:I_k \xrightarrow{\rm bijection} A$ such that $s\in A_i\cap A_j$ since $s$ is surjective it follows that for some $q\in I_k$, $f(q)= h(j)$ equivalently $(f(q),h(j))\in s$ but the definition of $A_j$ implies that no such ordered pair exists in $A_j$. 
$\square$
Lemma.(2). $ \cup_{i\in I_{k+1}}A_i \thicksim F$
Proof. Consider now the function $\mathcal{N}:\cup_{i\in I_{k+1}}A_i\to F$ defined as follows
$$\mathcal{N}(f) = f\cup\left\{(k+1,h(j))\right\}\ \operatorname{where}\ f\in A_j$$
With the above definition it  is not difficult to see that $\mathcal{N}$ is a is a bijection we leave the details to the reader.
$\square$
Using Lemma(1) in conjunction with (7) implies that $\cup_{i\in I_{k+1}A_i}$ is finite and 
$$|\cup_{i\in I_{k+1}}A_i| = \sum_{i=1}^{k+1}|A_i|$$
and by using (7) in conjunction with Lemma(2) we can deduce the following 
$$|F| = |\cup_{i\in I_{k+1}}A_i|$$
from the inductive hypothesis we know that $\forall i\in I_{k+1}(|A_i| = k!)$ consequently 
$$|F| = \sum_{i=1}^{k+1}|A_i| = \underbrace{k!+k!+\ \cdot\ \cdot\ \cdot\ +k!}_{k+1\operatorname{terms}} = (k+1)\cdot k! = (k+1)!$$
completing the inductive step.
$\blacksquare$
 A: I did not check every detail of the OP's proof, but it looks correct.
It appears that the OP wanted to prove the result by induction without using the usual and very simple rule of product argument. The following might be of interest here:
For $n \ge 1$ let $S_n$ denote the bijective mappings on $I_n = \{1,2,\dots,n\}$. 
Proposition 1: There is a canonical injection ${\iota_{n+1}^n}: S_n \to S_{n+1}$ sending the bijections in $S_n$ to the bijections in $S_{n+1}$ that keep $n+1$ fixed.
Proof: Exercise.
Let ${S_n}^{'} = {\iota_{n+1}^n}(S_n) \subset S_{n+1}$. 
Proposition 2: For every bijection $\sigma \in S_{n+1}$ not in ${S_n}^{'}$ there exists a unique $\sigma^{'} \in {S_n}^{'}$ and transposition of the form $(n+1 \, k) \,\text{with}\,k \le n$ such that
$\tag 1 \sigma =  \sigma^{'} \circ (n+1 \, k)$
Proof: Exercise.
For $1 \le k \le n$, let ${S'}^{\,k}_n = \{\sigma \in S_{n+1} \; | \; \sigma \text{ is in (1) form} \}$. For convenience, let ${S'}^{\,0}_n$ denote ${S_n}^{'}$. 
Proposition 3: The sets ${S'}^{\,k}_n$, with $0 \le k \le n$, partition $S_{n+1}$. Moreover, all the blocks is the partition can be put in bijective correspondence with ${S_n}^{'}$.
Proof: Exercise.
Using proposition 3 it is immediate that $|S_{n+1}| = |\text{ Union of  } (n+1) \text{ disjoint copies of } S_{n}|$.
Proposition 4: The sets $S_n$ are all finite with $|S_{n}| = n!$.
Proof: Hint: Use induction and proposition 3.
