Suppose $\mathbf{a}, \mathbf{b} \in \mathbb{R}^{n}$,
proof that $\left\langle\mathbf{a}^{\otimes d}, \mathbf{b}^{\otimes d}\right\rangle=\langle\mathbf{a}, \mathbf{b}\rangle^{d}$
where $\mathbf{a}^{\otimes d}=\underbrace{\mathbf{a} \otimes \mathbf{a} \otimes \ldots \otimes \mathbf{a}}_{d \text { times }} \quad$.
So when I try to prove this, I get from the left side
$\left\langle\mathbf{a}^{\otimes d}, \mathbf{b}^{\otimes d}\right\rangle=\sum_{i_{1}=1}^{n} \sum_{i_{2}=1}^{n} \cdots \sum_{i_{d}=1}^{n} a_{i_{1}} a_{i_{2}} \cdots a_{i_{d}} b_{i_{1}} b_{i_{2}} \cdots b_{i_{d}}$
and for the right side
$\langle\mathbf{a}, \mathbf{b}\rangle^{d}= (\sum_{i=1}^{n} a_{i} b_{i})^d$ .
Does anyone know how to te reach an equality from here? Thank you.