# Curiosity about a simple identity

Consider the well known formula

$$1^3 + 2^3 +\cdots+ n^3 = (1+\cdots+n)^2 , n \in N$$

Now suppose that for all $n \in N$ the identity is true :

$$1^k + 2^k +\cdots+ n^k = (1+\cdots+n)^{k-1}$$

with $k \in N$ fixed. Which is the possible values of $k$ ?

Someone could point me a reference?

Sorry for my english, it is not good

• $1 + 2^k = 3^k$ so $k=1$ – Will Jagy Sep 7 '14 at 2:57
The dominant term of the left hand side is $$\frac{1}{k+1} n^{k+1}.$$ The right hand side is $$\left( \frac{n^2 + n}{2} \right)^{k-1},$$ dominant term $$\frac{2}{2^k} n^{2k - 2}.$$
So $$k+1 = 2k - 2$$ and $$3 = k$$
Consider the case when $n = 2$. Then we have that $1 + 2^k = 3^{k-1}$. By Catalan's Conjecture (Mihăilescu's theorem), we have that $k = 3$ (the theorem basically states that the only two consecutive perfect powers are $2^3$ and $3^2$.)