Commutative ring product of elements in terms of sum of powers of sums The following fact is stated (but not proved) in Chapter 0 of Noll's Finite Dimensional Spaces (it's 07.21):

If $R$ is a commutative ring with unity and $r_1, ..., r_n$ elements of $R$ then
  $$
n!\text{ }r_1\cdots r_n=\sum_{k=1}^{n}(-1)^{n-k}\sum_{1\le i_1<\cdots<i_k\le n}(r_{i_1}+\cdots+r_{i_k})^n
$$

Now as it happens I (think I) can see a proof of this.
Start by showing that it holds when setting $r_{1}=X_{1}, ..., r_n=X_{n}$ in $\mathbb{Z}[X_{1}, ...,X_{n}]$. To do this, first show that the RHS is zero at $X_{i}=0$ so is divided by $X_{i}$, so is divided by the product $X_{1}\cdots X_{n}$ (by UFD-ness). Thus by considering degrees the RHS is a constant times this product. Setting $X_{i}=1$ then reveals this constant and gives the polynomial identity.
Then use the free-ness of $\mathbb{Z}[X_{1}, ...,X_{n}]$ to map these elements onto any $r_i$ in any commutative ring with 1, to deduce that the identity holds there.
However seeing as how this fact is stated without proof in the introductory chapter to a not particularly advanced textbook (a textbook that does not have knowledge of abstract algebra as a prerequisite), and appears in the company of some trivial facts like the binomial theorem, I was wondering if there was a more elementary proof of the identity?
 A: While I like your proof, here is one that does not use the fact that $\mathbb{Z}[X_1, \ldots, X_n]$ is a UFD.
As you pointed out, this identity holds "universally" by considering the elements $r_i$ as formal variables.  Doing this, it is easy to see that in the right hand side of the equation, the coefficient in front of $r_1\cdots r_n$ is $n!$.  What remains to be shown is that the coefficient in front of every other monomial is zero.
To do this, let's modify the notations a bit.  Let $I=\{1,\ldots n\}$.  Then the right hand side of the equation can be written as
$$ \sum_{k=1}^{n}(-1)^{n-k}\sum_{\stackrel{J\subseteq I}{|J|=k}}  (\sum_{\alpha\in J}r_\alpha)^n. $$
Now fix a monomial $r_{i_1}^{a_1}\cdots r_{i_l}^{a_l}$ with $\sum_{j=1}^{l}a_j = n$.  Fix $K=\{i_1, \ldots, i_l\}$.  The only terms in the above sum contributing to the coefficient in front of this monomial are those terms corresponding to a subset $J\subseteq I$ containing $K$.  Moreover, given such a subset $J$ containing $K$, the coefficient in front of $r_{i_1}^{a_1}\cdots r_{i_l}^{a_l}$ in $(\sum_{\alpha\in J}r_\alpha)^n$ is 
$$\binom{n}{a_1}\binom{n-a_1}{a_2}\cdots \binom{n-a_1-\ldots-a_{l-1}}{a_l}. $$
As such, it does not depend on $J$.  Finally, the number of subsets $J$ containing $K$ and such that $|J|=k$ is $\binom{n-l}{k-l}$ whenever $l\leq k \leq n$.
Therefore the coefficient in front of $r_{i_1}^{a_1}\cdots r_{i_k}^{a_k}$ in the sum is
$$ \sum_{k=l}^{n}(-1)^{n-k} \binom{n-l}{k-l} \binom{n}{a_1}\binom{n-a_1}{a_2}\cdots \binom{n-a_1-\ldots-a_{l-1}}{a_l}   $$
which is in turn equal to 
$$ \binom{n}{a_1}\binom{n-a_1}{a_2}\cdots \binom{n-a_1-\ldots-a_{l-1}}{a_l} \sum_{k=l}^{n}(-1)^{n-k} \binom{n-l}{k-l} $$
and, changing indices, to
$$ \binom{n}{a_1}\binom{n-a_1}{a_2}\cdots \binom{n-a_1-\ldots-a_{l-1}}{a_l} \sum_{k=0}^{n-l}(-1)^{n-l-k} \binom{n-l}{k}. $$
The alternating sum of the first $n-l$ binary coefficients is zero if $n-l>0$; thus the above expression is zero if $l\neq n$.  This finishes the proof.
