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The problem asks to prove the following equation:

\[
\frac{k! \cdot (k - 2)!}{(k - 1)! \cdot (k - 2)!} = k^2 - 2k
\]

### Proof:

1. First, simplify the left-hand side:
   \[
   \frac{k! \cdot (k - 2)!}{(k - 1)! \cdot (k - 2)!}
   \]
   The \((k - 2)!\) terms cancel out, leaving:
   \[
   \frac{k!}{(k - 1)!}
   \]

2. Now, recall the definition of factorial:
   \[
   k! = k \cdot (k - 1)!
   \]
   Substituting this into the fraction:
   \[
   \frac{k \cdot (k - 1)!}{(k - 1)!}
   \]
   The \((k - 1)!\) terms cancel out, simplifying to:
   \[
   k
   \]

3. Therefore, the left-hand side of the equation simplifies to \( k \).

4. Now consider the right-hand side of the equation:
   \[
   k^2 - 2k
   \]
   Factor the expression:
   \[
   k(k - 2)
   \]

5. Now compare both sides:
   \[
   k = k(k - 2)
   \]
   Clearly, this only holds for specific values of \( k \).

So the given equation holds under certain conditions, typically when \( k = 0 \) or another specific value depending on additional constraints.

Would you like further details or have any questions? Here are some follow-up questions:

1. Why does the factorial simplify the equation?
2. What does the expression \( k(k-2) \) represent in general?
3. Can you derive the general solution for \( k \)?
4. How can you apply this proof to other similar equations?
5. What happens when \( k = 2 \)?

**Tip**: Factorials cancel easily when they appear in both the numerator and denominator, simplifying complex expressions quickly!

Buktikan \(\frac{k! \cdot (k-2)!}{(k-1)! \cdot (k-2)!} = k^2 - 2k\)

Explore the proof of the algebraic identity \(\frac{k! \cdot (k-2)!}{(k-1)! \cdot (k-2)!} = k^2 - 2k\). This page covers factorial properties, simplification techniques, and algebraic identities suitable for university-level mathematics.

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