suppose k,n are integers ≥1 show that (k.n)! is divisible by (k!)^n

To demonstrate that \((k \cdot n)!\) is divisible by \((k!)^n\) for integers \(k\) and \(n\) where both are greater than or equal to 1, we can utilize the concept of combinatorial counting and the properties of factorials. Let’s break this down step by step.

Understanding Factorials

First, recall that the factorial of a number \(m\), denoted \(m!\), is the product of all positive integers from 1 to \(m\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). The factorial function grows very quickly as \(m\) increases.

Setting Up the Problem

We need to show that \((k \cdot n)!\) can be expressed in a way that clearly shows it contains \((k!)^n\) as a factor. To do this, we can think about how we can group the integers in \((k \cdot n)!\) into sets of \(k\).

Grouping Integers

Consider the integers from 1 to \(k \cdot n\). We can group these integers into \(n\) groups, each containing \(k\) consecutive integers. For instance, if \(k = 3\) and \(n = 2\), we would group the integers as follows:

• Group 1: {1, 2, 3}
• Group 2: {4, 5, 6}

In general, the groups can be represented as:

• Group 1: {1, 2, ..., k}
• Group 2: {k+1, k+2, ..., 2k}
• Group 3: {2k+1, 2k+2, ..., 3k}
• ...
• Group n: {(n-1)k+1, ..., nk}

Counting Arrangements

Now, within each group of \(k\), the number of ways to arrange those \(k\) items is given by \(k!\). Since there are \(n\) such groups, the total number of arrangements of all groups is \((k!)^n\).

Factorial Expansion

When we expand \((k \cdot n)!\), it represents the total number of ways to arrange \(k \cdot n\) distinct items. This can be thought of as arranging all the items from each of the \(n\) groups. Thus, we can express \((k \cdot n)!\) in terms of the arrangements of these groups:

Mathematically, we can express this as:

\((k \cdot n)! = (k!)^n \times \text{(number of ways to arrange the groups)}\)

Conclusion on Divisibility

Since \((k \cdot n)!\) can be factored into \((k!)^n\) multiplied by some integer (the number of ways to arrange the groups), it follows that \((k \cdot n)!\) is divisible by \((k!)^n\). This shows that the factorial of the product \(k \cdot n\) indeed contains \((k!)^n\) as a factor, confirming the divisibility.

In summary, by grouping the integers in \((k \cdot n)!\) into sets of \(k\) and recognizing that each group contributes a factor of \(k!\), we can conclude that \((k \cdot n)!\) is divisible by \((k!)^n\). This approach not only proves the statement but also illustrates the beauty of combinatorial reasoning in mathematics.