Let $(H,\mu,\eta,\Delta,\varepsilon)$ be a bialgebra with antipode $S$ which is cocommutative. On $\text{End}(H)$ we have the product $$f\ast g:=\mu\circ(f\otimes g)\circ\Delta\in\text{End}(H).$$
Using this (and the cocommutativity) I want to show that $\text{id}_H^{\ast n+1}$ is a coalgebra morphism. Any thoughts?
This just follows from a direct calculation. The map is $f:x\in H\mapsto x_1\cdots x_n\in H$.
Now \begin{align} \Delta(f(x))&=(x_1\cdots x_n)_1\otimes(x_1\cdots x_n)_2\\ &=(x_1)_1\cdots (x_n)_1\otimes (x_1)_2\cdots (x_n)_2\\ &= x_1x_3\cdots x_{2n-1}\otimes x_2x_4\cdots x_{2n} \end{align} and \begin{align} (f\otimes f)(\Delta(x)) &= f(x_1)\otimes f(x_2) \\ &= (x_1)_1\cdots (x_1)_n\otimes (x_2)_1\cdots (x_2)_n \\ &= x_1\cdots x_n\otimes x_{n+1}\cdots x_{2n} \end{align} and these two things are equal simply because of cocommutativity.