Find the closed form (as a function of $k$ and $n$) of the following series sum:

$$
\binom{k}{1} \sum_{i=1}^{n} i+\binom{k}{2} \sum_{i=1}^{n} i^{2}+\ldots+\binom{k}{k-1} \sum_{i=1}^{n} i^{k-1}
$$
$$
\binom{k}{1} \sum_{i=1}^{n} i+\binom{k}{2} \sum_{i=1}^{n} i^{2}+\ldots+\binom{k}{k-1} \sum_{i=1}^{n} i^{k-1}
$$

Multiply the terms under (1) by the binomial coefficient in front of them, and consider the $i$-th product from each group:

$$
\binom{k}{1} i, \quad\binom{k}{2} i^{2}, \ldots,\binom{k}{k-1} i^{k-1}
$$

These expressions remind us of the binomial theorem, and based on this, their sum can be easily determined:

$$
\begin{gathered}
\binom{k}{1} i+\binom{k}{2} i^{2}+\ldots+\binom{k}{k-1} i^{k-1}= \\
=\left[\binom{k}{0} \cdot 1+\binom{k}{1} i+\binom{k}{2} i^{2}+\ldots+\binom{k}{k-1} i^{k-1}+\binom{k}{k} i^{k}\right]- \\
-1-i^{k}=(i+1)^{k}-i^{k}-1
\end{gathered}
$$

These numbers need to be taken for $i=1,2, \ldots, n$ and summed:

$$
\sum_{i=1}^{n}\left[(i+1)^{k}-i^{k}-1\right]=\sum_{i=2}^{n+1} j^{k}-\sum_{i=1}^{n} i^{k}-n=(n+1)^{k}-1-n
$$

Thus, the value of the expression under (1) is $(n+1)^{k}-(n+1)$.

Remark. The last transformation in the solution can be stated as follows. Consider the difference between the $k$-th power of the number one greater than each natural number from 1 to $n$ and the $k$-th power of the number itself, and sum these differences. This sum can also be obtained by first summing the first terms of the differences and then subtracting the sum of the second terms. The sum of the first terms is nothing other than the sum of the $k$-th powers of the natural numbers from 2 to $(n+1)$, and the sum of the second terms is the sum of the $k$-th powers of the numbers from 1 to $n$. Therefore, everything cancels out except for the extreme terms: $(n+1)^{k}-1^{k}$.

We can also say that if we write the differences one after another:

$$
\left(2^{k}-1^{k}\right)+\left(3^{k}-2^{k}\right)+\ldots+\left((n+1)^{k}-n^{k}\right)
$$

and then change the order of the terms:

$$
\left(-1^{k}+2^{k}\right)+\left(-2^{k}+3^{k}\right)+\ldots+\left(-n^{k}+(n+1)^{k}\right)
$$

then the algebraic sum of the adjacent terms in each pair is 0, the entire sum collapses like an accordion, and only the difference of the extreme terms remains.