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Well, assume again that $V$ is a free $k$-module with base $X=x_i,i∈I$. One has to avoid the fact that $k\langle\langle X\rangle\rangle$ is NOT a Hopf algebra, be it with shuffle or concatenation except when $X=\emptyset$ because you have to take Sweedler's dual and cannot consider complete dual. A striking (but limited to free - finite or infinitely generated - $V$) proof of the first statement goes as follows

a) The Hopf algebra $k\langle X\rangle$ with concatenation as product and co-shuffle as coproduct is graded - in finite dimensions - over $\mathbb{N}^{(I)}$.

b) Then, the shuffle Hopf algebra is exactly the graded dual of it with the pairing given by
$$\langle x_{i_1}\otimes\ldots\otimes x_{i_n}\mid x_{j_1}\otimes\ldots\otimes x_{j_n}\rangle=\delta_{i_1,j_1}\ldots \delta_{i_n,j_n}$$ and 0 if $n\not=m$.

c) (For statement 2.) the antipode is just $S^*=S$ (the adjoint of $S$).

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Well, assume again that $V$ is a free $k$-module with base $X=x_i,i∈I$. One has to avoid the fact that $k\langle\langle X\rangle\rangle$ is NOT a Hopf algebra, be it with shuffle or concatenation except when $X=\emptyset$ because you have to take Sweedler's dual and cannot consider complete dual. A striking (but limited to free - finite or infinitely generated - $V$) proof of the first statement goes as follows

a) The Hopf algebra $k\langle X\rangle$ with concatenation as product and co-shuffle as coproduct is graded - in finite dimensions - over $\mathbb{N}^{(I)}$.

b) Then, the shuffle Hopf algebra is exactly the graded dual of it with the pairing given by
$$\langle x_{i_1}\otimes\ldots\otimes x_{i_n}\mid x_{j_1}\otimes\ldots\otimes x_{j_n}\rangle=\delta_{i_1,j_1}\ldots \delta_{i_n,j_n}$$ and 0 if $n\not=m$.