Understanding Fraleigh's proof of: Every finite integral domain is a field Here's how Fraleigh proves: Every finite integral domain is a field in his book:

Let 
          \begin{equation*}
        0, 1, a_1, \dots, a_n
    \end{equation*}
  be all the elements of the finite domain $D$. Now, consider 
          \begin{equation*}
        a1, aa_1, \dots, aa_n
    \end{equation*}
  Since the multiplicative cancellation laws hold in $D$, it means that each of $a1, aa_1, \dots, aa_n$ are distinct from each other since $aa_i = aa_j \implies a_i = a_j$. Also, since $D$ has no divisors of $0$, neither of $a1, aa_1, \dots, aa_n$ can be zero. Hence, $a1, aa_1, \dots, aa_n$ are elements $1, a_1, \dots, a_n$ in some order. So, either $a1 = 1 \implies a = 1$ or $aa_i = 1$ for some $i$. 

My addition: If $a = 1$, then the conditional in question is trivially satisfied and there is nothing to prove. So, without loss of generality, assume $aa_i = 1$. 

This shows that $a$ has a multiplicative inverse, $a_i$. $\square$

I have two questions: firstly, is my addition to the proof valid? Secondly, how does $D$ has no divisors of $0$ imply "neither of $a1, aa_1, \dots, aa_n$ can be zero" (in bold above). The definition of 0 divisors that Fraleigh has given is:

If $a$ and $b$ are two nonzero elements of a ring $R$ s.t. $ab = 0$, then $a$ and $b$ are divisors of 0. 

To conclude that "neither of $a1, aa_1, \dots, aa_n$ can be zero" from this definition, I think  we would need to know that the product of any two terms from $a1, aa_1, \dots, aa_n$ is zero but we don't know this. What am I missing? Thanks! 
 A: The whole point is to show that none of the products $a1,aa_1,\ldots,aa_n$ is $0$. Suppose that some $aa_k$ were $0$. We know that $a$ and $a_k$ are not $0$; if $aa_k$ were $0$, $a$ and $a_k$ would by definition be divisors of $0$, but we know that $D$ has no divisors of $0$. Thus, $aa_k$ cannot be $0$. The same argument shows that $a1$ cannot be $0$, though in that case it’s even easier, since $a1=a$, and we know that $a\ne 0$.
Your addition is correct but not really necessary: one would hope that the reader can be trusted to recognize that if $a=1$, we already know that it has a multiplicative inverse, so we’re really interested in the other cases.
A: Here is a quick little proof of this fact which I hope will help resolve our OP User31415 questions, and clarify his/her approach:
Let $D^\times$ be the set of non-zero elements of $D$:
$D^\times = \{0 \ne d \in D \}; \tag 1$
since $D$ is finite, 
$\vert D \vert < \infty, \tag 2$
so is $D^\times$; indeed,
$\vert D^\times \vert = \vert D \vert - 1; \tag 3$
now for 
$a, b \in D^\times \tag 4$
we have
$ab \in D^\times; \tag 5$
for otherwise
$ab = 0, \tag 6$
which contradicts the hypothesis that $D$ has no zero divisors; now (4)-(5) imply
$aD^\times = \{ab, \; b \in D^\times \} \subseteq D^\times; \tag 7$
thus we may consider the function
$\phi_a:D^\times \to D^\times \tag 8$
given by
$\phi_a(b) = ab, \; \forall b \in D^\times; \tag 9$
$\phi_a$ is injective, for with $a \ne 0$
$\phi_a(b) = \phi_a(c) \Longleftrightarrow ab = ac \Longleftrightarrow a(b - c) = 0 \Longleftrightarrow b = c; \tag{10}$
since $\phi_a$ is an injective function from $D^\times$ to itself, by virtue of (2)-(3) (that is, the finiteness of $\vert D^\times \vert$), it is also surjective; hence there is some $a' \in D^\times$ with
$\phi_a(a') = 1 \Longleftrightarrow aa' = 1, \tag{11}$
that is, $a$ has an inverse in $D^\times$.  Since every non-zero element of $D$ is invertible, $D$ is a field.
A: I don't know why you think that the product of any two terms from $a1,aa_1,\dots,aa_n$ would have to be zero.  That is not the case.
Fraleigh is obviously assuming that $a\ne0$.  Thus if any of $a1,aa_1,\dots,aa_n$ were zero, we would have a contradiction, since $D$ has no zero divisors.
A: Treating $0$ and $1$ separately from the rest of the elements of $D$ makes the proof unnecessarily complicated and seems to have produced some confusion. So here's a shorter and possibly less confusing version of the proof.
Apparently Fraleigh has already established the cancellation law for integral domains: If $ab=ac$ and $a\neq0$, then $b=c$. Now, to show that $D$ is a field, I'll consider an arbitrary non-zero element $a\in D$ and show that it has a multiplicative inverse. Multiplication by $a$ is a function $f:D\to D$, namely $f(x)=ax$ for all $x\in D$. Cancellation says exactly that this $f$  is one-to-one. From elementary set theory, we know that a one-to-one function from a finite set $D$ into itself must be onto. So $1$ is in the image of $f$, i.e., there is $x\in D$ with $f(x)=1$. That means $ax=1$, and so $x$ is the desired inverse of $a$.
