Theorem: For any integer n ≥ 2, the number n! (the factorial of n) is divisible by n.

Proof:

Step 1) We prove this by strong mathematical induction. The base case is n = 2. Here, 2! = 2, which is divisible by 2, so the statement is true for n = 2.

Step 2) Now assume the statement is true for some arbitrary integer k ≥ 2. That is, assume k! is divisible by k.

Step 3) We need to show that (k + 1)! is also divisible by (k + 1). By the definition of factorial, we have:

(k + 1)! = (k + 1) × k!

Step 4) Since k! is divisible by k (by assumption), there exists some integer m such that:

k! = mk

for some m.

Step 5) Substituting this into the expression for (k + 1)!, we have:

(k + 1)! = (k + 1) × (mk) = (k + 1)m × k

Since (k + 1)m and k are both integers, this shows that (k + 1)! is divisible by (k + 1).

Step 6) Therefore, by the principle of strong mathematical induction, the statement is true for all integers n ≥ 2.

That's the outline of a proof by strong mathematical induction showing that for any n ≥ 2, the factorial n! is divisible by n. Of course, this could be elaborated with more mathematical precision and rigour, but this captures the key steps of the inductive argument.
spam fails_task lang_mismatch pii not_appropriate hate_speech sexual_content quality toxicity humor helpfulness creativity violence