Proving that $V^{\otimes d} = U(\mathfrak g)v_0$ 
Let $d\leq n, V=\mathbb C^{n+1}$ with basis $\{v_1,\cdots,v_{n+1}\}$, $ \mathfrak g = \mathfrak{sl}_{n+1}$ and $v_0 = v_{i_1}\otimes \cdots \otimes v_{i_d}\in V^{\otimes d}$ such that $i_1,\cdots, i_d \in \{1,\cdots,n+1\}$ are all distinct. How do I prove that $U(\mathfrak g)v_0 = V^{\otimes d}$ ?

Some notation: denote by $x_i^\pm$ the standard generators of $\mathfrak g$ such that $[x_i^+,x_i^-] =h_i$ and by $x_\theta^\pm$ the root vectors associated with the longest root $\theta$.
Regarding the action of $\mathfrak g$ on $V$, recall that  $x_i^-v_j = \delta_{i,j}v_{j+1},
x_i^+v_j = \delta_{i,j-1}v_{j-1}$ and that $x_\theta^-v_j=\delta_{1,j}v_{n+1},x_\theta^+v_j= \delta_{n+1,j}v_1.$ Also, $\mathfrak g$ acts on $V^{\otimes d}$ by $x(v_1\otimes \cdots \otimes v_d) = \sum_k v_1\otimes \cdots \otimes x.v_k\otimes \cdots \otimes v_d$.
Toughts so far: It suffices to show that all vectors of the form $v_{j_1}\otimes \cdots \otimes v_{j_d} \in U(\mathfrak g)v_0.$ I tried to use induction over $d$. It is valid for $d=1$, since $v_0 = v_i \in V$ and $v_j = x_{j-1}^- \cdots x_i^-v_0$ (supposing that $i<j$).
Since $x(v_{i_1}\otimes \cdots \otimes v_{i_d}) = (x(v_{i_1}\otimes \cdots \otimes v_{i_{d-1}})\otimes v_{i_d} + v_{i_1}\otimes \cdots \otimes v_{i_{d-1}}\otimes x .v_{i_d}$, using the induction hyphotesis, there exists a $x\in U(\mathfrak g)$ such that
$$ x(v_{i_1}\otimes \cdots \otimes v_{i_d})  = v_{j_1}\otimes \cdots \otimes v_{j_{d-1}}\otimes v_{i_d}+v_{i_1}\otimes \cdots \otimes v_{i_{d-1}}\otimes x.v_{i_d}.$$
Here comes the problem. Altough there exists $x' \in U(\mathfrak g)$ such that $x'.v_{i_d} = v_{j_d},$ it might be the case that some ${j_k} = i_d, 1\leq k \leq d-1,$ so applying $x'$ to both sides of the above equality will make the right hand side a sum of $v_{j_1}\otimes \cdots \otimes v_{j_d}$ with some other summands. If I could show that those other summands lie in the set $U(\mathfrak g)v_0$, its done. But that does not seem easy to argue.
Any insight on how to prove this? Thanks.
 A: Too long for a comment but here's a start:
Since $v_1, v_2, \dots, v_{n+1}$ are linearly independent we can always find $x \in \mathfrak{sl}_{n+1}$ such that $x(v_j) = 0$ for $j \ne k$ and $x(v_k) = \sum_{\ell \ne k} a_\ell v_\ell$ for any constants $a_\ell$ we want (the trace zero condition is why there is no $a_k$ term). If you apply this to a vector $v_{i_1} \otimes \dots \otimes v_{i_d}$, if there is a term $v_{i_m} = v_k$ it will change it to $\sum_{\ell \ne k} a_\ell v_\ell$ but leave the rest of the terms the same.  Therefore applying the operator $a_\ell\cdot1 + x \in U(\mathfrak{g})$ we can change the term $v_k$ in the tensor to any other vector $v \in \mathbb{C}^{n+1}$ without changing the other tensor terms.
The idea is to then iterate this sort of construction to turn the starting vector into any other pure tensor $w_1 \otimes \dots \otimes w_d$ changing one term at a time.  I think a little care needs to be taken for when $w_1, \dots, w_d$ span a lower dimensional subspace, but that shouldn't be too hard (or maybe induction can handle that).
