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Bernoulli stated sum of series of powers as:

Bernoulli stated sum of series of powers as

LINK to the image source (Power Sum)

I had a doubt in the given formula in the picture! What if $n < p$ i.e. $1^4 + 2^4 + 3^4$ here $n = 3$ and $p = 4$ so outer summation runs ($i = 1$ to $p$) from $1$ to $4$. And we have to find $(n-i)!$ [factorial]. Won't it lead to factorial of negative number when $i > n$ ?

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There is no factorial in the formula, there are binomial coefficients. If you look up the definition carefully, you will note that for nonnegative integers $n,k$ we have

$${n \choose k} := \begin{cases} \frac{n!}{k!(n-k)!}, & k \leq n \\ 0 & else\end{cases}.$$

Thus, you will not run into trouble as there will be no factorials in your case.

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  • $\begingroup$ As a sidenote: since $\binom{n}{k}$ is the number of ways of choosing $k$ objects from a set of $n$ objects, how many ways are there of choosing $4$ objects from a set of $3$ objects? It must be that $\binom{n}{k}=0$ when $k>n$. $\endgroup$ – Jam Nov 17 '17 at 13:09

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