Discrete maths/set theory question 
*22. Suppose $\lvert A \rvert=n$, and let $F=\{f \mid \,f \text{ is a one-to-one, onto function from } I_n \text{ to } A\}.$ 
  $I_n=\{i∈Z^+ \mid i≤n\}$. $\lvert A\rvert$ is the cardinality of A.
(a) Prove that $F$ is finite, and $\lvert F\rvert=n!$. (Hint: Use induction on $n$.)
(b) Let $L = \{ R \mid \, R \text{ is a total order on } A \}$. Prove that $F \sim L$, and therefore $\lvert L\rvert= n!$.
(c) Five people are to sit in a row of five seats. In how many ways can they be seated?

From  http://users.metu.edu.tr/serge/courses/111-2011/textbook-math111.pdf p.g. 314.  I've seen answers and don't fully understand.  My main issue is part (a) but might as well give answer to all.  Can anyone give a clearer write up of the answer.  Alternative methods also welcome.  
UPDATE:  I need help with the below answer.  I don't how to prove that $h$ is one-to-one based on the suggestion.  Why is $(f(i), g(i))$ an element of $h(f)$ and not $h(g)$?

 A: To prove (a) using induction over $n$:
Base: $n = 1$. So we have: $|A| = 1$, or $A = \{ a_1 \}$ for some object $a_1$, and we have $I_1 = \{ 1 \}$. Then there is only $1$ function $f$ possible from $I_1$ to $A$, namely $f(1) = a_1$. Thus, $|F| = 1 = 1!$. Check!
Step: Assume (inductive hypothesis) that $|F| = k!$ for some $k$.  Now let's use that to show $|F|= (k+1)!$ in the case of $k+1$.
OK, so we have $|A|=k+1$, i.e. $A = \{ a_1, a_2, ..., a_k, a_{k+1} \}$. We also have $I_{k+1} = \{ 1,2,3,..., k, k+1 \}$. Now, how many one-to-one and onto functions are there from $I_{k+1}$ to $A$?  Well, let's do this by first mapping $k+1$ to an element in $A$, and then mapping all the other numbers $1$ through $k$. 
So, what can $f(k+1)$ be?  Given that this is the first object we map, all options are open, so it can be mapped to any element in $A$, so there are $k+1$ options: $f(k+1) = a_1$, or $f(k+1) = a_2$, or ... or  $f(k+1) = a_{k+1}$. Now, once we have fixed $f(k+1)$, there are $k$ elements left in $A$ that all the others numbers $1$ through $k$ need to be mapped to in a one-to-one and onto way. Or, maybe a little more cleanly: if $f(k+1) = a_i$, then the set $A' = A \setminus \{ a_i \}$ will have $k$ elements, i.e. $|A'|=k$. So, we can apply our inductive hypothesis: there will be $|F|=k!$ ways to map elements $1$ through $k$ (i.e. $I_k$) to $A'$.
So, since there are $k+1$ ways to map $k+1$ to an element in $A$, and $k!$ ways to map all the other numbers $1$ through $k$ to the rest of $A$ in a one-to-one and onto way, there are $(k+1)*k! = (k+1)!$ ways to map $I_{k+1}$ to $A$ in a one-to-one and onto way. Check!
A: For part ($b$):
Let $X := \{j \in I_n \mid f(j) \neq g(j)\}$; observe that $X \neq \varnothing$ (because $f \neq g$ by assumption), so it has minimum (being a non-empty subset of $\mathbb N$). Define, then, $i := \min X$ (in particular $i \in X$); we have that $(f(i), g(i)) \in h(f)$ if and only if (using the definition of $h(f)$) $f^{-1}(f(i)) \leq f^{-1}(g(i))$, that is (since $f$ is bijective) if and only if $i \leq f^{-1}(g(i))$. To see that this holds, we show that $f^{-1}(g(i)) \in X$: in facts, $f(f^{-1}(g(i))) = g(i) \neq g(f^{-1}(g(i)))$, where the last inequality holds because if it were $g(i) = g(f^{-1}(g(i)))$, since $g$ is injective we would get $i = f^{-1}(g(i))$, and since $f$ is a bijective function we would obtain $f(i) = f(f^{-1}(g(i))) = g(i)$, which contradicts the fact that $i \in X$. Thus $f(f^{-1}(g(i))) \neq g(f^{-1}(g(i)))$, and so $f^{-1}(g(i)) \in X$; then $i \leq f^{-1}(g(i))$ being $i$ the minimum of $X$ (more precisely, we have $i < f^{-1}(g(i))$ since we just proved that equality leads to a contradiction), and so we deduce that $(f(i), g(i)) \in h(f)$.
In a similar way, we can prove that $(f(i), g(i)) \notin h(g)$: first, we have $(f(i), g(i)) \notin h(g) \iff g^{-1}(f(i)) > g^{-1}(g(i)) \iff g^{-1}(f(i)) > i$, so we can prove that $g^{-1}(f(i)) \in X \setminus \{i\}$. This is true because $i = g^{-1}(f(i))$ leads again to the contradiction $g(i) = f(i)$, so $i \neq g^{-1}(f(i))$; then $g(g^{-1}(f(i))) = f(i) \neq f(g^{-1}(f(i)))$. Thus $g^{-1}(f(i)) \in X \setminus \{i\}$, which means that $g^{-1}(f(i)) > i$, and so we conclude $(f(i), g(i)) \notin h(g)$.   
