Symmetrization is the unique $k$-tensor

Question

$\newcommand{\Sym}[1]{\operatorname{Sym}{#1}}$

Let $V$ be a $n$-dim real vector space with dual space $V^*$. Let $\alpha$ be a covariant $k$-tensor, i.e., $\alpha \in T^k(V^*) \equiv (V^*)^{\otimes k}$. Then how would you show that the symmetrization $\Sym{\alpha}$ of $\alpha$ is the unique symmetric $k$-tensor such that $$ \boxed{\Sym{\alpha} (v,...,v) =\alpha(v,...,v), \qquad v\in V} $$

Note the symmetrization is defined by $$ \Sym{\alpha} (v_1,...,v_k) = \frac{1}{k!} \sum_{\sigma\in S_k} \alpha (v_{\sigma_1},...,v_{\sigma_k}) $$ Where $S_k$ is the symmetric group of $k$ and $T^k(V^*)$ is identified as the space of multi-linear real functionals on $V^k$.

EDIT: The key seems to be proving the following fact $$ \boxed{k!(v_1\cdots v_k) = \sum_{l=0}^k (-1)^l \sum_{|J|=l,J\subseteq \{1,...,k\}}\left( \sum_{i\in\{1,...,k\}-J} v_i \right)^k} $$ where the $|J|$ is the number of elements in set $J$. I used $v_1\cdots v_k$ to denote $\beta(v_1,..., v_k)$ where $\beta$ is a symmetric $k$-tensor. Similarly, I used $v^k$ to denote $\beta(v,...,v)$.

E.g. when $k=3$, we have $$ 3!(abc)=(a+b+c)^3-(a+b)^3-(a+c)^3-(b+c)^3+a^3+b^3+c^3 $$ However I'm having troubles proving this identity for general $k$.

EDIT 2: I just learned that the formula in the first EDIT refers to the polarization formula, which can be found in this post

Answer 1

The idea is the following:

$k=2$

In this case $V^*\otimes V^*$ is the space of bilinear forms on $V$ in which you know that it is possibile identifies each symmetric bilinear forms by its quadratic form:

For each $\beta\in V^*\otimes V^*$ you have that

$\beta(v,w)=\frac{1}{2}(\beta(v+w,v+w)-\beta(v-w,v-w))$

so you have that if $\beta$ is a symmetric tensor such that

$\beta(v,v)=\alpha(v,v)$ then $\beta=\alpha$