Polarization formula.

This is true, here's a proof.

I'm going to use the polynomial notation $\Phi\left(v_{1},\ldots,v_{n}\right)=v_{1}\cdots v_{n}$ - note that the multilinearity and symmetry of $\Phi$ means that manipulating these like polynomials (i.e. commuting elements, distributing ``multiplication'') is completely legitimate. Let the RHS of your proposed equation be $\frac{1}{n!}F\left(n\right)$.

Using the multinomial expansion, we have $$ F\left(n\right)=\sum_{k=1}^{n}\left(-1\right)^{n-k}f\left(n,k\right) $$ where $$ f\left(n,k\right)=\sum_{1\le j_{1}<\cdots<j_{k}\le n}\ \sum_{l_{1}+\cdots+l_{k}=n}{n \choose l_{1},\ldots,l_{k}}v_{j_{1}}^{l_{1}}\cdots v_{j_{k}}^{l_{k}}. $$ Let's try to compute the coefficient of $v_{j_{1}}^{l_{1}}\cdots v_{j_{k}}^{l_{k}}$ in $F\left(n\right)$. The most obvious contribution is from $f\left(n,k\right)$, which gives $\left(-1\right)^{n-k}{n \choose l_{1},\ldots,l_{k}}.$ But there are more contributions: for every $K>k$ we have terms where $K-k$ of the $l$s are zero. The contribution from $f\left(n,K\right)$ is $$ \left(-1\right)^{n-K}\sum\left\{ {n \choose l_{1},\ldots,l_{k},0,0,\ldots}:j_{k+1},\ldots,j_{K}\textrm{ distinct from }j_{1},\ldots,j_{k}\right\} . $$ All we need to do to compute this is count the number of choices of $K-k$ of the $n-k$ remaining indices, so we get $$ \left(-1\right)^{n-K}{n \choose l_{1},\ldots,l_{k}}{n-k \choose K-k}. $$ The coefficient of $v_{j_{1}}^{l_{1}}\cdots v_{j_{k}}^{l_{k}}$ in $F\left(n\right)$ is thus $$ \sum_{K=k}^{n}\left(-1\right)^{n-K}{n \choose l_{1},\ldots,l_{k}}{n-k \choose K-k}=\sum_{K=k}^{n}\left(-1\right)^{n-K}\frac{n!}{l_{1}!\cdots l_{k}!}{n-k \choose K-k}. $$ We want to show that this is $n!$ when $k=n,l_{j}=1$ and zero otherwise. The first case is easy - there is only a single term in the sum and all of $n,k,K$ are just $n$, so it falls out immediately. Let's try the zero case. Factoring out the $K$-independent terms gives $$ \left(-1\right)^{n}\frac{n!}{l_{1}!\cdots l_{k}!}\sum_{K=k}^{n}\left(-1\right)^{K}{n-k \choose K-k}. $$ Making a change of variables $j=K-k$ turns the sum to $$ \left(-1\right)^{k}\sum_{j=0}^{n-k}\left(-1\right)^{j}{n-k \choose j}. $$ This is the alternating sum of the binomial coefficients, which vanishes as required.

SKETCH OF THE PROOF : Your big sums are always sums of (sums of sums of) terms of the form $\Phi(v_{k_1},v_{k_2},\ldots ,v_{k_n})$ for some uples of indices $(k_1,k_2, \ldots ,k_n)$. Thanks to the symmetry of $\Phi$, we can always rearrange and put the uple in increasing order. You are then left with a simpler sum with less terms, where exactly one term is multilinear (the term $\Phi(v_1,v_2, \ldots ,v_n)$) and all the others are not. So this is a Rambo-like situation of one against a hundred. But fortunately for us, the combinatorial property (let us call it $P$) that for any finite set $X$, the sum $\sum_{B \subseteq X}(-1)^{|B|}$ is zero except when $X$ is empty, allows us to show that all the non-multilinear terms have zero coefficient in the sum.

THE DETAILS : Given a uple $(k_1,\ldots ,k_n)$, denote by $\rho(k_1,k_2,\ldots,k_n)$ the rearranged uple according to increasing order. (Thus, $\rho(1,3,2)=(1,2,3)$).

It is more convenient here to view uples as functions, so we shall speak of $u$ and $\rho u$ where $u$ and $\rho u$ are maps $\lbrace 1,2, \ldots, n\rbrace \to \lbrace 1,2, \ldots, n\rbrace$ and $\rho u$ is increasing. Also, we put $\psi(f)=\Phi(v_{f(1)},\ldots,v_{f(n)})$.

For an arbitrary increasing uple $i$, denote by $w(i)$ the number of uples $j$ satisfying $\rho(j)=i$. For any $A \subseteq \lbrace 1,2, \ldots ,n \rbrace$, denote by $I(A)$ the set of all increasing maps $\lbrace 1,2, \ldots ,n \rbrace \to A$. Also, let $I=I(\lbrace 1,2, \ldots, n \rbrace)$ and $V(f)=\lbrace A \subseteq \lbrace 1,2, \ldots ,n \rbrace | f\in I(A) \rbrace$ . Note that if $K(f)=\lbrace 1,2, \ldots ,n \rbrace \setminus Im(f)$, then there is a natural bijection between ${\cal P}(K(f))$ and $V(f)$, given by $B \mapsto Im(f) \cup B$.

Let $$ \lambda (A)=\phi\bigg(\sum_{a\in A}v_a\bigg) \tag{2} $$

Then, expanding $\lambda(A)$ completely shows that

$$ \lambda (A)=\sum_{f\in I(A)} w(f) \psi(f) \tag{3} $$

Then, the RHS (call it $\Phi''$) of the desired equality can be rewritten as

$$ \begin{eqnarray} \Phi'' &=& \sum_{A\subseteq \lbrace 1,2, \ldots ,n \rbrace}(-1)^{n-|A|}\lambda(A)\\ &=& \sum_{A\subseteq \lbrace 1,2, \ldots ,n \rbrace}(-1)^{n-|A|}\sum_{f\in I(A)} w(f) \psi(f) \\ &=& \sum_{f\in I}w(f)\psi(f)\sum_{A\in V(f)}(-1)^{n-|A|} \\ &=& \sum_{f\in I}w(f)\psi(f)\sum_{B\subseteq K(f)}(-1)^{n-|Im(f)|+|B|} \\ &=& w({\mathsf{id}})\psi(\mathsf{id}) \ \text{by property } P. \\ &=& n! \Phi(v_1,v_2, \ldots ,v_n) \end{eqnarray} $$

which concludes the proof.