We are tasked with proving the given combinatorial identity for integers $n$ and $k$ where $1 \leq k \leq n$. The identity is:

$k \binom{n}{k} = n \binom{n-1}{k-1}.$

Proof:

1. Expression for Binomial Coefficient:

   Recall the definition of the binomial coefficient: $\binom{n}{k} = \frac{n!}{k!(n-k)!}.$

2. Left-hand Side:

   The left-hand side is: $k \binom{n}{k} = k \cdot \frac{n!}{k!(n-k)!}.$ Simplify: $k \binom{n}{k} = \frac{k \cdot n!}{k!(n-k)!}.$ Since $k! = k \cdot (k-1)!$, this simplifies further: $k \binom{n}{k} = \frac{n!}{(k-1)!(n-k)!}.$

3. Right-hand Side:

   The right-hand side is: $n \binom{n-1}{k-1}.$ Using the binomial coefficient definition, we expand: $\binom{n-1}{k-1} = \frac{(n-1)!}{(k-1)!((n-1)-(k-1))!} = \frac{(n-1)!}{(k-1)!(n-k)!}.$ Multiply by $n$: $n \binom{n-1}{k-1} = n \cdot \frac{(n-1)!}{(k-1)!(n-k)!}.$ Since $n \cdot (n-1)! = n!$, this becomes: $n \binom{n-1}{k-1} = \frac{n!}{(k-1)!(n-k)!}.$

4. Conclusion:

   Comparing the simplified expressions for the left-hand side and the right-hand side: $k \binom{n}{k} = \frac{n!}{(k-1)!(n-k)!} = n \binom{n-1}{k-1}.$

   Thus, the identity is proven.

Would you like further clarification or examples?

Here are five related questions to explore:

1. What is the combinatorial interpretation of this identity?
2. Can you derive a similar identity involving $\binom{n}{k+1}$?
3. How is this identity used in Pascal’s triangle?
4. Can you extend this proof to the generalization of multinomial coefficients?
5. How does this identity help in probability and statistics?

Tip: Always verify combinatorial identities using small numbers as examples to reinforce understanding!