How is the highest order term in the summation of kth powers generalized?

Question

More specifically, I read that the highest order term for $\sum_{i=1}^n i^k$ is $\frac{1}{k+1}$$n^k$ because "Gauss’s trick is that $k + 1$-tuples with a single largest component have that component in one of $k + 1$ places" (http://www.math.caltech.edu/~nets/cranks.pdf). Is there an intuitive, combinatorial explanation for this?

Answer 1

$(n+1)^{k+1}-n^{k+1}=(k+1)n^k +F (n,k)$

[F(n,k) has highest power 'k-1' of n]

$(n)^{k+1}-(n-1)^{k+1}=(k+1)(n-1)^k+F(n-1,k)$

........

........(writing till 1)

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adding them all -> $(n)^{k+1} -1 = (k+1)(\sum_{r=1}^n r^k) +F(n,k) + F(n-1,k) ..............$

(obviously coefficient of highest power of n which is 'k+1' is $\frac{1}{k+1}$)