Supplementary exercise 51, chapter 3 from 'A walk through combinatorics' 4th edition I'm selfstudying 'A walk through combinatorics' by Miklós Bóna. This book has some supplementary exercises at the end of each chapter, no solution provided. I'm trying exercise 51, chapter 3.
Problemstatement:

A store has $n$ different products for sale. Each of them has a different price that is at least one dollar, at most $n$ dollars, and is a whole dollar. A customer only has the time to inspect $k$ different products. After doing so, she buys the product that has the lowest price among the $k$ products she inspected. Prove that on average she will pay $\frac{n+1}{k+1}$ dollars.

Attempt at solution: The customer pays:

*

*1 dollar if the product with price 1 dollar is the cheapest among the $k$ inspected products. This is possible in $\binom{n-1}{k-1}$ ways, since we need to pick $k-1$ products with prices in $\{2, \ldots, n\}$.

*2 dollars if the product with price 2 dollars is the cheapest among the $k$ inspected products. This is possible in $\binom{n-2}{k-1}$ ways, since we need to pick $k-1$ products with prices in $\{3, \ldots, n\}$.

*$\ldots$

*$n-k+1$ dollar if the product with price $n-k+1$ is the cheapest among the $k$ inspected products. There are only $k-1$ products which are more expensive, so this is possible in $\binom{n-(n-k+1)}{k-1} = \binom{k-1}{k-1}$ ways.

This implies that the customer has to pay, on average
$$\frac{1}{\binom{n}{k}}\cdot\left(1\cdot \binom{n-1}{k-1} + 2 \cdot \binom{n-2}{k-1} + \ldots + (n-k+1) \cdot \binom{k-1}{k-1}\right).$$
This is where I'm stuck: I want to show that the sum in brackets equals $\binom{n+1}{k+1}$ (guess based on what we need to prove, checked with some values of $n,k$. This seems to be correct).
Question: How to prove that
$$1\cdot \binom{n-1}{k-1} + 2 \cdot \binom{n-2}{k-1} + \ldots + (n-k+1) \cdot \binom{k-1}{k-1} = \binom{n+1}{k+1},$$
a hint would be appreciated. I tried to give a combinatorial proof, but failed.

A small hint for alternative:

*

*List all possible $(k+1)$ - subsets of $\{0,1,...,n\}$

*Remove the smallest element from each subset

You'll end up with a list of all possible $k$ - subsets of $\{1,...,n\}$ but with each subset repeated as many times as its smallest element.
How many entries are in the list? $\binom{n+1}{k+1}$
