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The problem of finding expansions of monomials, binomials etc. is classical and there is a lot of beautiful solutions have been found already, the most prominent examples are

Binomial Theorem, Wikipedia.

For every $(a,b,n)\in\mathbb{N}$ $$(a+b)^n=\sum_{k=0}^n \binom{n}{k}a^k b^{n-k},$$ and for monomials could be written as $$m^n=\sum_{k=0}^{n}\sum_{j=0}^{k}\binom{n}{k}\binom{k}{j}(-1)^{k-j}m^j.$$

It also follows an identity in binomials to Falling Factorial power $$(x+y)^\underline{n} = \sum_{k=0}^n \binom{n}{k} x^\underline{k} y^\underline{n-k}$$

Multinomial Theorem, Wikipedia. (Partial Case)

For every $(m,n)\in\mathbb{N}$ $$m^{n}=\sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}},$$ where ${n \choose k_{1},k_{2},\ldots ,k_{m}}$ is Multinomial coefficient.

Faulhaber's Formula, arXiv, pp. 9-10.

For every $(m,n)\in\mathbb{N}$ $$n^{2m-1} = \sum_{k=1}^{m} (2k-1)!T(2m,2k) \binom{n+k-1}{2k-1},$$ where $T(2m,2k)$ are Central Factorial Numbers.

Worpitzky Identity, https://eudml.org/doc/148532.

For every $(m,n)\in\mathbb{N}$ $$m^n = \sum_{k=0}^{n} E_{n,k} \binom{m+k}{n},$$ where $E_{n,k}$ are Eulerian Numbers.

Identity between Stirling numbers of the second kind and falling factorial

For every $(x,n)\in\mathbb{N}$ $$x^n=\sum _{{k=0}}^{n}\left\{{n \atop k}\right\}(x)_{k},$$ where $\left\{{n \atop k}\right\}$ are Stirling numbers of the second kind and $(x)_{k}$ is Falling factorial.

Also the one good example can be found at Wolfram Mathworld, so-called

MacMillan Double Binomial Sum, Wolfram MathWorld, "Power", eq. 12.

For every $(x,n)\in\mathbb{N}$ $$x^n=\sum_{k=1}^{n}\sum_{j=1}^{k}(-1)^{k-j}\binom{k}{j}\binom{x}{k}j^n.$$

Proof of MacMillan identity is discussed here.

Result from question 2669237. $$m^n=\sum_{\mathbf{k}\in {\cal K}_m^n}\frac{m!n!}{\displaystyle\prod_{i\ge0}(i!)^{k_i}k_i!},$$ see the link in title for the explanations.

Ascending Bases and Exponents in Pascal's Triangle, Cut the Knot. $$(n+1)^j=\frac{j!\prod_{k=0}^j\binom{n+k}{k}}{\prod_{k=0}^{j-1}\binom{n+1+k}{k}}$$

Pascal's identity, arXiv. $$(n+1)^{k+1}-1=\sum_{m=1}^{n}[(m+1)^{k+1}-m^{k+1}]=\sum_{p=0}^{k}\binom{k+1}{p}(1^p+2^p+\cdots+n^p),$$


Question: Are there any other power identites, which not in the list above ?

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  • $\begingroup$ I think questions 1 and 2 should be separate posts. $\endgroup$ – Mike Earnest Apr 6 at 0:26
  • $\begingroup$ Good idea and its done $\endgroup$ – Petro Kolosov Apr 6 at 16:24
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Question 1: Certainly there are other identities, for example the following one: $$ m^n=\sum_{k_i\ge0}^{\sum_i k_i=m\atop\sum_i ik_i=n}\frac{m!n!}{\displaystyle\prod_{i\ge0}(i!)^{k_i}k_i!}. $$

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