How to prove that the discriminant is a symmetric function? My approach:
We know that the discriminant for a polynomial $x^n - s_1x^n-1 + s_2x^n-2 + ...$ where $s_i$ is the elementary symmetric function. 
is $(u_1-u_2)^2 (u_1-u_3)^2 ... (u_{n-1} - u_{n})^2$
So we want to show that this expression does not vary under interchanging terms. 
We see that these terms consist of pairwise combinations of $u_1, ...u_n$
so it cannot vary when we change the order of arguments.
I don't think that's formal enough though. Is it the right track?
 A: Because the set of transpositions generates the whole symmetric group $S_n$, it suffices to show that swapping two $u_i$'s will not change the value in question.  So if the value is fixed under any transposition, then it's also fixed under any permutation of the $u_i$'s.  
Suppose we want to interchange $u_j$ and $u_k$.  Notice that for $u_j$, there exists a term in the product that looks like $(u_j - u_m)^2$ or $(u_m - u_j)^2$ for all $1 \leq m \neq j \leq n$, and likewise for $u_k$.  So exchanging them won't change the value of the overall product since we will still have a term that looks like $(u_j - u_m)$ or $(u_m - u_j)$ for each $m \neq j$ (and likewise for $u_k$).  The only potential worry is a sign change due to the subtractions, but this is no problem because $(u_j - u_m)^2 = (u_m - u_j)^2$.
A: It is clear that the discriminant can be written as 
$$
\prod_{\{i,j\}\in T}(u_i-u_j)^2,
$$ 
where $T$ is the set of subsets of cardinal two of $\{1,\dots,n\}$, and it is equally clear that this product is invariant under permutation.
