Space of symmetric $n$-tensors spanned by elements of form $\underbrace{v \otimes \ldots \otimes v}_{n}$? Let $k$ be a field of characteristic $0$, and let $V$ be a finite dimensional vector space over $k$. Consider the space of symmetric $n$-tensors,$$S^nV = (V \otimes V \otimes \ldots \otimes V)^{S_n},$$where $\otimes = \otimes_k$ and the symmetric group $S_n$ acts on the $n$-fold tensor product$$V^{\otimes n} = V \otimes V \otimes \ldots \otimes V$$by permuting the factors. How do I see that $S^nV$ is spanned by elements of the form$$\underbrace{v \otimes \ldots \otimes v}_{n}, \quad v \in V?$$I've seen some proofs online, but they're way too terse or long and not easily understandable to a beginner like me. Can someone give me a straight to the point sketch of a proof that's clear?
 A: Since you have proofs available, perhaps the most useful thing is for you to see the general pattern.  Of course, the statement is trivial for $n = 1$. 
Let $e_1,\dots,e_d$ form a basis of $V$, so that the vectors of the form $e_{i_1} \vee \cdots \vee e_{i_n}$ space $V$. For $n = 2$, we see that for $i \neq j$, we have
$$
e_i \vee e_j = (e_i + e_j) \otimes (e_i + e_j) - e_i \otimes e_i - e_j \otimes e_j
$$
so that we hit every element of that spanning set.  Similarly, for $n = 3$, we have
$$
e_i \vee e_j \vee e_j + e_i \vee e_i \vee e_j = \\
(e_i + e_j) \otimes (e_i + e_j) \otimes (e_i + e_j) - e_i \otimes e_i \otimes e_i - e_j \otimes e_j \otimes e_j\\
e_i \vee e_j \vee e_j - e_i \vee e_i \vee e_j = \\
(e_i - e_j) \otimes (e_i - e_j) \otimes (e_i - e_j) - e_i \otimes e_i \otimes e_i - e_j \otimes e_j \otimes e_j
$$
(note: your coefficients may vary) At this point, we see that we have all elements of the form $u_1 \vee u_2 \vee u_2$ in the span.
From there, we can use the idea from $n = 2$:
$$
e_i \vee e_j \vee e_k = 
e_i \vee (e_j + e_k) \vee (e_j + e_k) - e_i \vee e_j \vee e_j - e_i \vee e_k \vee e_k
$$
Perhaps now you can see that pattern which will persist.
