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 $(x_{j})^{-1}$?
Remember that cancellation holds in domains. That is, if $c \neq 0$, then $ac = bc$ implies $a=b$. So, given $x$, consider $x, x^2, x^3,......$. Out of finiteness there would be a repetition sometime: $x^n  = x^m$ for some $n &gt;m$. Then, by cancellation, $x^{n-m} =1$, and $x$ has an inverse.
Why to find the inverse of $xj$
Can't it be solved this way
$(x1.x2...xk) - xj = 0$
Then $xj(x1..xj-1.xj+1..xk  -1)=0$
As $xj$ is not a $0$ and $R$ is integral
We get $x1..xj-1.xj+1..xk = 1$
Choosing an arbitrary $xm$
$Xm(x1..xm-1.xm+1..xj-1.xj+1..xk)=1$
So $xm$ has an inverse