$K$-dimensional generalization of identity $[a_1^{n+1},a_2^{n+1}]= (a_1+a_2) [ a_1^n,a_2^n ]-a_1a_2 [a_1^{n-1}, a_2^{n-1}]$

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Is there a generalization $K$-dimensional generalization of the follows 2-dimensional identity \begin{align} [a_1^{n+1},a_2^{n+1}]= (a_1+a_2) [ a_1^n,a_2^n ]-a_1a_2 [a_1^{n-1}, a_2^{n-1}] \end{align} where $a_1$ and $a_2$ are positive reals. Where $[a_1,a_2]$ is a vector of two elements.
In other words, can we similarly re-write \begin{align} [a_1^{n+1}, a_2^{n+1},..., a_K^{n+1}] \end{align} similarly.

Also, does this identity have a name?

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When we write the $2$-dim identity as $$[a_1^{n+2}, a_2^{n+2}] - (a_1 + a_2)[a_1^{n+1}, a_2^{n+1}] + a_1a_2[a_1^{n}, a_2^{n}] = 0$$ then the $K$-dim generalization would be $$\sum_{j = 0}^K (-1)^{K-j} e_j(a_1, \ldots, a_K)[a_1^{N+K-j}, \ldots, a_{K}^{N+K-j} ] = 0$$ where $e_j(a_1, \ldots, a_K)$ is the $j$th elementary symmetric polynomial.

Thus is just saying that $a_1, \ldots, a_K$ are roots of the polynomial $$\begin{align*} f(x) & = \sum_{j = 0}^K (-1)^{K-j} e_j(a_1, \ldots, a_K) x^{K-j} \\ & = \prod_{i=1}^K (x - a_i) \end{align*}$$

See Vieta's formulas.