Lemma 2

If $x, y\in A$ and $n\in \mathbf{N}$,

(ad$nx$! )$^ny=\sum(-1)^qxp^p$! $yxq^q$! = $\sum(-1)^qx^{(p)}yx^{(q)}$. (11)

$p+q=np+q=n$

Indeed, if we denote by $L_x$ and $R_x$ the maps $z \rightarrow xz$ and $z \rightarrow zx$ from A to A, we have, since $L_x$ and $R_x$ commute,

$n1$!(ad $x$)$^n=n1$!$(L_x-R_x)^n=\sum_{p+q=n}(-1)^qp1$! $L^p_xq1$! $R^q_x$.