I'm studying the coproduct $\Delta$ defined on a tensor algebra $T(V)$ and its action on tensor products of elements from a vector space $V$. The coproduct is given by $\Delta(v) = v \boxtimes 1 + 1 \boxtimes v$ for any $v \in V$, and extended as an algebra homomorphism over $T(V)$ (through the universal property of $T(V)$). I've encountered the following formula that expresses $\Delta$ applied to a tensor product of $m$ elements:
$$ \Delta(v_1)\otimes\cdots\otimes\Delta(v_m) = \sum_{p=0}^m \; \sum_{\sigma\in\mathrm{Sh}(p,m-p)} \; \left(v_{\sigma(1)}\otimes\dots\otimes v_{\sigma(p)}\right) \boxtimes \left(v_{\sigma(p+1)}\otimes\dots\otimes v_{\sigma(m)}\right) $$
Inductive Hypothesis: Assume the formula holds for a tensor product of $m-1$ elements.
Inductive Step: To prove the formula for $m$ elements, we consider the action of $\Delta$ on the tensor product of $m$ elements and then apply the inductive hypothesis.
Starting from the inductive hypothesis, the application of $\Delta(v_m)$ yields:
$$ \sum_{p=0}^{m-1} \sum_{\sigma\in\mathrm{Sh}(p,m-1-p)} \left[ (v_1 \otimes \cdots \otimes v_p) \otimes v_m \boxtimes (v_{p+1} \otimes \cdots \otimes v_{m-1}) \otimes 1 + (v_1 \otimes \cdots \otimes v_p) \otimes 1 \boxtimes (v_{p+1} \otimes \cdots \otimes v_{m-1}) \otimes v_m \right] $$
We wish to show this equals:
$$ \sum_{p=0}^m \; \sum_{\sigma\in\mathrm{Sh}(p,m-p)} \; \left(v_{\sigma(1)}\otimes\dots\otimes v_{\sigma(p)}\right) \boxtimes \left(v_{\sigma(p+1)}\otimes\dots\otimes v_{\sigma(m)}\right) $$
I find it hard to see why this last step works. How does one rigorously justify this equality?
Maybe it has something to do with the inductive definition of the shuffle product?