The reason is Schur–Weyl duality. $\newcommand{\GL}{\operatorname{GL}}$

The subspace $W = \langle \forall v \in V \, \mid \, v \otimes v \otimes \cdots \otimes v\rangle$ forms a $\GL(V)$ subrepresentation of $\bigotimes^kV$ if we allow $\GL(V)$ to act diagonally on tensors.

If we consider the dual action, which is the symmetric group $S_k$ permuting tensor factors, we see that all of the generators of $W$ have the symmetry type of the trivial representation of $S_k$ since they are invariant under these permutations. It follows that the $\GL(V)$ subrepresentation $W$ is contained within $\mbox{Sym}^kV \subset \bigotimes^kV$.

However, by Schur-Weyl duality, the symmetric tensors form an irreducible representation of $\GL(V)$ — the subrepresentation $W$ is either $0$ or all of $\mbox{Sym}^kV$.

It isn't $0$, so every symmetric tensor $s$ can be written

$$s = \alpha_0 \cdot v_0 \otimes v_0 \otimes \cdots \otimes v_0 +
\alpha_1 \cdot v_1 \otimes v_1 \otimes \cdots \otimes v_1 +
\cdots
\alpha_l \cdot v_l \otimes v_l \otimes \cdots \otimes v_l$$

for some suitable choice of $v_i$ and $\alpha_i$.

In other words, the $k^{\rm th}$ powers of the elements of $V$ span $\mbox{Sym}^k V$. It follows that knowing a symmetric multilinear form on the $k^{\rm th}$ powers is enough to determine the form.