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sothea hoeung
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Prove that (n!)! is divisible by (n!)^(n-1)!
To prove that (n) is divisible by (n)^(n-1), we can use the definition of divisibility which states that if (a) is divisible by (b), then there exists an integer (k) such that (a) = (k)(b). In this case, we can let (a) = (n) and (b) = (n)^(n-1). Then, we can rewrite (n)^(n-1) as (n^n)/(n), which simplifies to (n)^(n-2). Therefore, we can see that (n)^(n-2) is a factor of (n) and thus (n) is divisible by (n)^(n-1).
Yes, there are several methods that can be used to prove that (n) is divisible by (n)^(n-1). One method is to use mathematical induction, where we first prove that the statement is true for a base case (such as (n) = 1) and then show that if the statement is true for (n), it is also true for (n+1). Another method is to use the definition of divisibility and factorization to show that (n)^(n-1) is a factor of (n).
Yes, let's take (n) = 5. Then, (n)^(n-1) = 5^(5-1) = 5^4 = 625. We can see that 625 is a factor of 5, as 5/625 = 1. Therefore, we can say that 5 is divisible by 5^4.
Proving that (n) is divisible by (n)^(n-1) is important because it allows us to understand the relationship between a number and its powers. It also helps us to better understand the concept of divisibility and how to manipulate and simplify expressions involving exponents. Additionally, this proof can be applied in various mathematical contexts, such as in number theory and algebra.
Yes, the proof for (n) being divisible by (n)^(n-1) can be extended to show that (n) is divisible by (n)^m, where m is any positive integer. This can be done by using the same logic as in the original proof, but instead of using (n)^(n-1), we use (n)^(m-1). This shows that (n)^(m-1) is a factor of (n), and therefore (n) is divisible by (n)^m.