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For 1, we WLOG assume that $V$ is a finite free $k$-module (because all we have to prove are some identities involving finitely many elements of $V$; now we can see these elements as images of a map from a finite free $k$-module $W$, and by functoriality it is thus enough to prove these identities in $W$). Then, we have $V^{\ast\ast}\cong V$, and we notice that the dual of our above Hopf algebra (we don't know that it is a Hopf algebra yet, but at least it has the right signature) is the tensor Hopf algebra of $V^{\ast}$, for which Hopf algebraicity is much easier to show.

For 1, we WLOG assume that $V$ is a finite free $k$-module (because all we have to prove are some identities involving finitely many elements of $V$; now we can see these elements as images of a map from a finite free $k$-module $W$, and by functoriality it is thus enough to prove these identities in $W$). Then, we have $V^{\ast\ast}\cong V$, and we notice that the graded dual of our above graded Hopf algebra (we don't know that it is a graded Hopf algebra yet, but at least it has the right signature) is the tensor Hopf algebra of $V^{\ast}$, for which Hopf algebraicity is much easier to show. (Note that this only works with the graded dual, not with the standard dual, because $TW$ is free but not finite free.)

One reason why I am thinking that there are simpler proof is that the similar statements for the tensor Hopf algebra (this is another Hopf algebra with underlying $k$-module $TV$; it has the same counit and unit map as the shuffle Hopf algebra, but the multiplication is the standard tensor algebra multiplication, and the comultiplication is the so-called shuffle comultiplication) are significantly easier to prove. In particular, 2 holds verbatim for the tensor Hopf algebra, but the proof is almost trivial (since $v_1\otimes v_2\otimes ...\otimes v_n$ equals $v_1\cdot v_2\cdot ...\cdot v_n$ in the tensor Hopf algebra).

As far as I understand, we can identify the $k$-algebra $\left(TV,\mathrm{shf},\eta\right)$ with the symmetric part of the tensor algebra $TV$ if $k$ is a field of characteristic $0$ (or, more generally, a $\mathbb Q$-algebra), but I am not sure about what this does to the coalgebra structure, and even if it respects the coalgebra structure, this would only give us a proof of the above statements for characteristic $0$, and we would have to do some additional work to generalize it (although that is the easy part).

One reason why I am thinking that there are simpler proofs is that the similar statements for the tensor Hopf algebra (this is another Hopf algebra with underlying $k$-module $TV$; it has the same counit and unit map as the shuffle Hopf algebra, but the multiplication is the standard tensor algebra multiplication, and the comultiplication is the so-called shuffle comultiplication) are significantly easier to prove. In particular, 2 holds verbatim for the tensor Hopf algebra, but the proof is almost trivial (since $v_1\otimes v_2\otimes ...\otimes v_n$ equals $v_1\cdot v_2\cdot ...\cdot v_n$ in the tensor Hopf algebra).

What would Grothendieck do? Is there a good functorial interpretation, i. e., is the algebraic group induced by the shuffle Hopf algebra (since it is commutative) of any significance?