Let $R[X,\, Y,\, Z]$ denote the ring of the polynomials in noncommuting indeterminates $X,\, Y,\, Z$ over $R$. Define an automorphism $\sigma$ on $R[X,\, Y,\, Z]$ by:
$\sigma[F(X,\, Y,\, Z)]=F(X+1,\, Y,\, Z)$
and a $\sigma$-derivation $\Delta=\sigma-I$:
$\Delta[F(X,\, Y,\, Z)]=F(X+1,\, Y,\, Z)-F(X,\, Y,\, Z)$.
For two polynomials $F$ and $G$ in $R[X,\, Y,\, Z]$ we have (which is easy to show by definition of $\Delta$):
\begin{equation} \Delta[F+G]=\Delta[F]+\Delta[G] \,\,\, (1) \end{equation}
\begin{equation} \Delta[FG]=(\Delta[F])(\sigma[G])+F(\Delta[G]) \,\,\, (2) \end{equation}
and induction gives the Leibniz Formula
\begin{equation} \Delta^{n}[FG]=\sum_{r=0}^{n}\begin{pmatrix}n\\ r \end{pmatrix}(\Delta^{r}[F])(\sigma^{r}\Delta^{n-r}[G]) \end{equation}
I would like to prove this formula by induction. It is easily verified that this is true for $n=1$ and supposing that this is true for $n$ we get for $n+1$
$\Delta^{n+1}[FG]=\Delta(\Delta^{n}[FG])=\Delta\Bigl(\sum_{r=0}^{n}\begin{pmatrix}n\\ r \end{pmatrix}(\Delta^{r}[F])(\sigma^{r}\Delta^{n-r}[G])\Bigr)=\sum_{r=0}^{n}\Bigl(\Delta\begin{pmatrix}n\\ r \end{pmatrix}(\Delta^{r}[F])(\sigma^{r}\Delta^{n-r}[G])\Bigr)=\sum_{r=0}^{n}\Bigl(\begin{pmatrix}n\\ r \end{pmatrix}\Delta(\Delta^{r}[F])(\sigma^{r}\Delta^{n-r}[G])\Bigr) $
Could anyone help me to finish this proof, please? I'm stucked at this point and I can't move forward.
I've done few more steps of these induction.
$\sum_{r=0}^{n}\Bigl(\begin{pmatrix}n\\ r \end{pmatrix}\Delta(\Delta^{r}[F])(\sigma^{r}\Delta^{n-r}[G])=\sum_{r=0}^{n}\begin{pmatrix}n\\ r \end{pmatrix}\Delta^{r+1}[F]\sigma^{r+1}\Delta^{n-r}[G]+\sum_{r=0}^{n}\begin{pmatrix}n\\ r \end{pmatrix}[F]\sigma^{r}\Delta^{n-r-1}[G]=[F]\Delta^{n+1}[G]+\sum_{r=1}^{n}\begin{pmatrix}n\\ r \end{pmatrix}[F]\sigma^{r}\Delta^{n-r-1}[G]+\sum_{r=0}^{n-1}\begin{pmatrix}n\\ r \end{pmatrix}\Delta^{r+1}[F]\sigma^{r+1}\Delta^{n-r}[G]+\Delta^{n+1}[F]\sigma^{n+1}[G] $
And I am stucked again at these point. How can I finish these proof?
Thank you.