I don't get a part of Bernstein's / Deligne-Morgan's proof of Poincaré-Birkhoff-Witt Question: I am talking about the proof given on pages 50-52 of Pierre Deligne, Pavel Etingof, Daniel S. Freed, Lisa C. Jeffrey, David Kazhdan, John W. Morgan, David R. Morrison, and Edward Witten (editors), Quantum Fields and Strings: A Course for Mathematicians, Volume 1, AMS 1999 (on google books and in the usual internet sources).
The problem is easily described: In the middle of page 52, the authors say "and (1.3.7.7) gives that [...]". But I don't see how (1.3.7.7) gives the equation that follows.
Sidenotes: The proof was rather readable and well-written up to that point, so I assume the blindness is on my side. If anyone wishes to read the proof (or reprint the book ;) ), here are a few minor mistakes to watch out for:


*

*On page 51, $\left[xy\right]$ should be $\left[x,y\right]$ in "while the second term $\frac12\left[xy\right]$ is antisymmetric".

*On page 51, in the definition of the map $\left\lbrace x_1,...,x_{n+1}\right\rbrace$, all three terms on the right hand side should end with $x_{n+1}$ rather than $x_n$.

*On page 52, in the first formula of this page, the commutators $\left[x\left[y,z\right]\right]$ and $\left[z\left[x,y\right]\right]$ should be $\left[x,\left[y,z\right]\right]$ and $\left[z,\left[x,y\right]\right]$ instead.

*On page 52, in the middle of this page, "and the $\left\lbrace x_1,...,x_n\right\rbrace$ vanish" should probably be "and the $\left\lbrace x_1,...,x_{n+1}\right\rbrace$ vanish".

*On page 52, in the middle of this page, "1.3.7.4" should be "(1.3.7.4)".
 A: Here is an explanation Pavel Etingof has given to me in email. Thanks Pavel!
Every $\sigma\in S_{n}$ satisfies
$\sum\limits_{i}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast\left[  x,y_{\sigma i}\right]  \ast\cdots\ast y_{\sigma n}$
$=n\left[  x,y_{\sigma1}\right]  \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}$
$+\sum\limits_{i>1}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast\left[  x,y_{\sigma i}\right]  \ast\cdots\ast y_{\sigma n}$.
But since
$\sum\limits_{i>1}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast\underbrace{\displaystyle \left[
x,y_{\sigma i}\right]  }_{\displaystyle\substack{\displaystyle =x\ast y_{\sigma i}-y_{\sigma i}\ast
x\\\displaystyle \text{(since the inclusion of }\mathfrak{L}\\\displaystyle \text{into }\operatorname*{Sym}
\nolimits^{\ast}\mathfrak{L}\text{ is a morphism}\\\displaystyle \text{of Lie algebras)}
}}\ast\cdots\ast y_{\sigma n}$
$=\sum\limits_{i>1}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast\underbrace{\displaystyle \left(
x\ast y_{\sigma i}-y_{\sigma i}\ast x\right)  \ast\cdots\ast y_{\sigma n}
}_{\substack{\displaystyle =x\ast y_{\sigma i}\ast\cdots\ast y_{\sigma n}-y_{\sigma i}\ast
x\ast\cdots\ast y_{\sigma n}\\\displaystyle \text{(by the induction hypothesis, since }i>1\text{)}}}$
$=\sum\limits_{i>1}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast x\ast y_{\sigma
i}\ast\cdots\ast y_{\sigma n}$
$-\sum\limits_{i>1}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast
x\ast\cdots\ast y_{\sigma n}$
$= \sum\limits_{i>0}\left(  n-\left(  i+1\right)  +1\right)  \underbrace{\displaystyle y_{\sigma1}\ast\cdots\ast x\ast y_{\sigma
\left(i+1\right)}\ast\cdots\ast y_{\sigma n}}_{\displaystyle =y_{\sigma1}\ast\cdots\ast
y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}}$
$-\sum\limits_{i>1}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast
x\ast\cdots\ast y_{\sigma n}$ (here we substituted $i+1$ for $i$ in the first sum)
$=\sum\limits_{i>0}\left(  n-\left(  i+1\right)  +1\right)  y_{\sigma1}\ast\cdots\ast
y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$
$-\sum\limits_{i>1}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast
x\ast\cdots\ast y_{\sigma n}$
$=\left(  n-1\right)  y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\sum\limits
_{i>1}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$,
this becomes
$\sum\limits_{i}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast\left[  x,y_{\sigma
i}\right]  \ast\cdots\ast y_{\sigma n}$
$=n\left[  x,y_{\sigma1}\right]  \ast y_{\sigma2}\ast\cdots\ast y_{\sigma
n}+\left(  n-1\right)  y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$
$-\sum\limits_{i>1}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast
y_{\sigma n}$
$=n\left[  x,y_{\sigma1}\right]  \ast y_{\sigma2}\ast\cdots\ast y_{\sigma
n}+ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$
$-y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\sum\limits_{i>1}y_{\sigma1}\ast
\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$
$=n\left[  x,y_{\sigma1}\right]  \ast y_{\sigma2}\ast\cdots\ast y_{\sigma
n}+ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$
$-\sum\limits_{i>0}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma
n}$.
Thus, (1.3.7.7) rewrites as
$\dfrac{1}{n!}x\ast\sum\limits_{\sigma}y_{\sigma1}\ast\cdots\ast y_{\sigma n}=\left(
\text{symmetrized product of }x,y_{1},...,y_{n}\right)  $
$+\dfrac{1}{\left(  n+1\right)  !}\sum\limits_{\sigma}n\left[  x,y_{\sigma1}\right]
\ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left(  n+1\right)
!}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$
$-\dfrac{1}{\left(  n+1\right)  !}\sum\limits_{\sigma}\sum\limits_{i>0}y_{\sigma1}\ast\cdots\ast y_{\sigma
i}\ast x\ast\cdots\ast y_{\sigma n}$.
Since
$\left(  \text{symmetrized product of }x,y_{1},...,y_{n}\right)  $
$=\dfrac{1}{\left(  n+1\right)  !}\sum\limits_{\sigma}\left(  \sum\limits_{i>0}y_{\sigma1}
\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}+x\ast y_{\sigma
1}\ast\cdots\ast y_{\sigma n}\right)  $,
this simplifies to
$\dfrac{1}{n!}x\ast\sum\limits_{\sigma}y_{\sigma1}\ast\cdots\ast y_{\sigma n}
=\dfrac{1}{\left(  n+1\right)  !}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast
y_{\sigma n}$
$+\dfrac{1}{\left(  n+1\right)  !}\sum\limits_{\sigma}n\left[  x,y_{\sigma1}\right]
\ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left(  n+1\right)
!}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$.
Thus
$\dfrac{1}{\left(  n+1\right)  !}\sum\limits_{\sigma}n\left[  x,y_{\sigma1}\right]
\ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left(  n+1\right)
!}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$
$=\dfrac{1}{n!}x\ast\sum\limits_{\sigma}y_{\sigma1}\ast\cdots\ast y_{\sigma n}
-\dfrac{1}{\left(  n+1\right)  !}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast
y_{\sigma n}$
$=\underbrace{\displaystyle \left(  \dfrac{1}{n!}-\dfrac{1}{\left(  n+1\right)  !}\right)
}_{\displaystyle =\dfrac{n}{\left(  n+1\right)  !}}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast
\cdots\ast y_{\sigma n}=\dfrac{n}{\left(  n+1\right)  !}\sum\limits_{\sigma}x\ast
y_{\sigma1}\ast\cdots\ast y_{\sigma n}$.
Divide this by $\dfrac{n}{\left(  n+1\right)  !}$ to obtain
$\sum\limits_{\sigma}\left[  x,y_{\sigma1}\right]  \ast y_{\sigma2}\ast\cdots\ast
y_{\sigma n}+\sum\limits_{\sigma}y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$
$=\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n}$.
In other words,
$0=\sum\limits_{\sigma}\left(  x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n}
-y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\left[  x,y_{\sigma1}\right]
\ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}\right)  $
$=\sum\limits_{\sigma}\left\lbrace  x,y_{\sigma1},...,y_{\sigma n}\right\rbrace  =\left(
n-1\right)  !\sum\limits_{i}\left\lbrace  x,y_{i},y_{1},...,\widehat{y_{i}},...,y_{n}
\right\rbrace  $
(here we used that $\left\lbrace  x_{1},...,x_{n+1}\right\rbrace  $ is symmetric in the
last $n-1$ variables, so that every $\sigma\in S_{n}$ satisfies $\left\lbrace
x,y_{\sigma1},...,y_{\sigma n}\right\rbrace  =\left\lbrace  x,y_{i},y_{1}
,...,\widehat{y_{i}},...,y_{n}\right\rbrace  $ for $i=\sigma1$).
Thus, $\sum\limits_{i}\left\lbrace  x,y_{i},y_{1},...,\widehat{y_{i}},...,y_{n}\right\rbrace
=0$, qed.
