I was studying some definition about norm and distance on vector spaces, and I crossed this theorem:
Let $q:W \times W \to \mathbb{K}$ be a fixed inner product on vector space $W$, then there is a unique inner product on $Sym^d W $ (homogenous polynomials of degree $d$ over $W$), such that $$Q:Sym^d W \times Sym^d W \to \mathbb{K}, \ \ Q(f^d,g^d)=q(f,g)^d.$$ I know that the difinition of $Q(f,g)$ is given by: (we know there is coresspondance between homogenous polynomials of degree $d$ over $W$ and symmetric tensor of order $d$)
Consider a symmetric tensor $f\in Sym^dW$, it may be seen in coordinate systems as $$f = \sum_{|\alpha|=d}{d \choose \alpha}f_{\alpha} x^{\alpha},$$ where $\alpha = (\alpha_1, \dots , \alpha_n) \in \mathbb{Z}^n_{\geq 0}$, $|\alpha| \ \ \colon= \alpha_1 + \cdots + \alpha_n$, $x^{\alpha} \ \ \colon= x_1^{\alpha_1} \dots x_n^{\alpha_n}$ and ${d \choose \alpha} \ \ \colon= \frac{d!}{{\alpha_1}! \dots{\alpha_n}!}$ is multinomial coefficient. Then for $f, g \in Sym^d W$, the inner product is defined as \begin{align*} Q(\sum_{|\alpha|=d}{d \choose \alpha} f_{\alpha} x^{\alpha},\sum_{|\alpha|=d}{d \choose \alpha} g_{\alpha}x^{\alpha}) \ \ \colon= \sum_{|\alpha|=d}{d \choose \alpha}f_{\alpha}g_{\alpha} = \sum_{|\alpha|=d} \frac{d!}{{\alpha_1}! \dots{\alpha_n}!}f_{{\alpha_1} \dots{\alpha_n}}g_{{\alpha_1} \dots{\alpha_n}}. \end{align*} This definition holds the inner products' conditions (it is easy to check). Now I am stuck on proving $Q(f^d,g^d)=q(f,g)^d$.
Any comment and guidance are highly appreciated.