Let $T\_1,T\_2\in \cal{A}$ with $\cal{A}$ is an algebra.



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> Let $n\_1,n\_2\in \mathbb{N}$. Is it true that
>  $$[T\_1^{n\_1},T\_2^{n\_2}]=\displaystyle\sum\_{\substack{\alpha+\alpha'=n\_1-1 \\ \beta +\beta'=n\_2-1}}T\_1^{\alpha}T\_2^{\beta}[T\_1,T\_2]T\_2^{\alpha'}T\_1^{\beta'}?$$
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If the formula is false, is it possible to establish a relationship between $[T\_1^{n\_1},T\_2^{n\_2}]$ and $[T\_1,T\_2]$?



Commutator of the power of two elements
First try the case $n\_2 = 1$:


$$\eqalign{[T\_1^{n\_1}, T\_2] &= \sum\_{j=0}^{n\_1-1} (T\_1^{n\_1-j} T\_2 T\_1^{j} - T\_1^{n\_1-j-1} T\_2 T\_1^{j+1})\cr
&= \sum\_{j=0}^{n\_1-1} T\_1^{n\_1-j-1} [T\_1,T\_2] T\_1^{j}}$$
Similarly,
$$ [T\_1, T\_2^{n\_2}] = \sum\_{k=0}^{n\_2-1} T\_2^{n\_2-k-1} [T\_1,T\_2] T\_2^{k}$$
Substituting $T\_2^{n\_2}$ for $T\_2$ in the first formula and using the second,
$$[T\_1^{n\_1}, T\_2^{n\_2}] = \sum\_{j=0}^{n\_1-1} \sum\_{k=0}^{n\_2-1} T\_1^{n\_1-j-1} T\_2^{n\_2-k-1} [T\_1,T\_2] T\_2^k T\_1^j $$


