Let $X\subset\mathbb{C}^d$ be a (finite, non-empty) set of unit vectors. A standard way to define $X$ being a spherical $t$-design, is to impose it saturates the Welch bounds for all $k\le t$. Following the notation in https://arxiv.org/abs/quant-ph/0502031, this means $$\frac{1}{|X|^2} \sum_{x,y\in X}|\langle x|y\rangle|^{2k} = \frac{1}{\binom{d+k-1}{k} }.\tag1$$ Analogous definitions are used in https://arxiv.org/abs/1510.02767 and other related papers. I think this generalises to weighted designs by simply replacing the LHS with $\sum_{x,y\in X}p_x p_y \langle x|y\rangle|^{2k}$ for some probability distribution $(p_x)_x$.
However, I'm pretty sure I've seen an alternative definition used in the literature based on the identity $$S_k \equiv \sum_{x\in X} p_x |x\rangle\!\langle x|^{\otimes k} = \frac{\Pi_{\rm sym} }{\binom{d+k-1}{k}}, \qquad k=1,...,t,\tag2$$ where $\Pi_{\rm sym}$ is the projection on the completely symmetric subspace of $(\mathbb{C}^d)^{\otimes k}$.
It's pretty easy to see that (2) implies (1), as taking the inner product of the operator with itself, we get $$\sum_{x,y\in X}p_x p_y |\langle x|y\rangle|^{2k} = \operatorname{tr}(S_k^2) \equiv \langle S_k,S_k\rangle = \frac{\operatorname{tr}(\Pi_{\rm sym})}{\binom{d+k-1}{k}^2}= \frac{1}{\binom{d+k-1}{k} }.$$ How is the other direction proved? Equivalently, what's a reference showing this?