One identity in Lie algebras

Question

Let $L$ be a (non-restricted) Lie algebra over a field of prime characteristic $p,$ $UL$ be its universal enveloping algebra and $a_1,\dots, a_p \in L$ (the number of elements is equal to the characteristic). I can prove that the element $\sum_{\sigma \in S_p} a_{\sigma(1)}\cdot{}\dots{}\cdot a_{\sigma(p)}$ of $UL$ lies in $L.$ But I want to have an evident formula for this element on the language of commutators.

Question: Is it true that $$\sum_{\sigma \in S_p} a_{\sigma(1)}\cdot{}\dots{}\cdot a_{\sigma(p)} = \sum_{\sigma\in S_p\ :\ \sigma(1)=1} [a_{1},a_{\sigma(2)},\dots,a_{\sigma(p)}]$$ ?

Here $S_p$ denotes the symmetric group and $[x_1,\dots,x_n]=[[x_1,\dots,x_{n-1}],x_n].$ I can prove this formula for $p=2,3.$

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Answer: Yes, it is true. This formula can be found here: http://msp.org/agt/2006/6-5/agt-v6-n5-p08-p.pdf on page 2252, line 6 (without prove). Jie Wu explained me a prove.

Answer 1

If you expand the commutator by the possible $2^{n-1}$ sign choices you get $$[a_1,a_2,\dots,a_n]=\sum_{r=1}^{n-1}(-1)^r\sum_{k_1> k_2>\cdots> k_r>1}a_{k_1}a_{k_2}\cdots a_{k_r}W(k_1,k_2,\dots,k_r)$$ where the expression $W(k_1,k_2,\dots,k_r)$ stands for the word $a_1a_2\cdots a_n$ with the letters $a_{k_i}$ removed. Now this tells us that when you expand the sum $$\sum_{\sigma\in S_p, \sigma(1)=1}[a_{\sigma(1)},a_{\sigma(2)},\dots,a_{\sigma(p)}],$$ a word where $a_1$ is in the $r$-th position appears with coefficient $(-1)^r\binom{p-1}{r}$. Wison's theorem tells you that $(-1)^r\binom{p-1}{r}\equiv 1\pmod{p}$ and your equality follows.