Why is a finite integral domain always field? This is how I'm approaching it: let $R$ be a finite integral domain and I'm trying to show every element in $R$ has an inverse:


*

*let $R-\{0\}=\{x_1,x_2,\ldots,x_k\}$, 

*then as $R$ is closed under multiplication $\prod_{n=1}^k\ x_i=x_j$, 

*therefore by canceling $x_j$ we get $x_1x_2\cdots x_{j-1}x_{j+1}\cdots x_k=1 $, 

*by commuting any of these elements to the front we find an inverse for first term, e.g. for $x_m$ we have $x_m(x_1\cdots x_{m-1}x_{m+1}\cdots x_{j-1}x_{j+1}\cdots x_k)=1$, where $(x_m)^{-1}=x_1\cdots x_{m-1}x_{m+1}\cdots x_{j-1}x_{j+1}\cdots x_k$.


As far as I can see this is correct, so we have found inverses for all $x_i\in R$ apart from $x_j$ if I am right so far. How would we find $(x_{j})^{-1}$?

In fact, we can go a bit farther, and say that if $R$ is a finite commutative ring that has elements that are not zero-divisors, then $R$ has an identity.  Furthermore, every nonzero element of $R$ is either a unit or a zero-divisor.
To see why, pick $a\in R\setminus\{0\}$ with $a$ not a zero-divisor.  As $R$ is finite, the set $\{a,a^2,a^3,...\}$ must also be finite, whence there exist $m,n\in \mathbb{N}$ with $m<n$ and $a^m=a^n$.  
We will now show that $a^{n-m}$ serves as an identity for $R$.  Pick any $x\in R$.  Then $a^m=a^n$ implies $a^mx=a^nx$, whence $a^m(a^{n-m}x-x)=0$.  Now, since $a$ is not a zero divisor, it is clear that $a^m$ is not a zero-divisor.  Thus, the only way we can have $a^m(a^{n-m}x-x)=0$ is if $a^{n-m}x-x=0$ or $a^{n-m}x=x$.  Therefore $a^{n-m}=1_R$, and $R$ has an identity.
In fact, the proof of why any nonzero zero-divisor is a unit essentially follows from the same argument as above (letting $x=1$ now that we know that $R$ has an identity):  if $a\in R\setminus\{0\}$ is not a zero-divisor, then there exist $0<m<n$ with $a^m=a^n\,\Rightarrow\,a^m(a^{n-m}-1)=0\,\Rightarrow\,a^{n-m}=1$ (since, again, if $a$ is not a zero-divisor, then neither can $a^m$ be a zero-divisor).  Therefore, every nonzero element of $R$ is either a zero-divisor or a unit.
From here, it directly follows that every finite integral domain is a field, since integral domains have no zero-divisors.
