Prove by induction that n! > n^2 for all n greater than or equal to 4.

This is a fairly typical example of a question from the Further Maths syllabus.

We wish to demonstrate that for all integers n greater than or equal to 4, n! > n² .

We briefly recap that n! is the product of all numbers from 1 up to n. For example 4! = 4x3x2x1 = 24.

 As with any induction question, there are four steps that we will work through in turn. They are the base case, the induction hypothesis, the inductive step, and finally the conclusion.

We begin with the base case. Here we select the minimum integer satisfying the conditions, in this case n=4, and demonstrate that the inequality holds true. Clearly 4! = 24 > 4² = 16

 The second step is to assume our induction hypothesis.
For this, we simply assume that the inequality holds for n=k, for any valid integer k.
That is, we assume k! > k².
We will now use this in step three to show that IF the inequality holds true for n=k, then it also holds for n=k+1.

For step three we consider the left hand side of the inequality, for n=k+1, and show that it is indeed larger than the right hand side for the same n:

 (k+1)! = (k+1) x k x (k-1) x (k-2) x.....x 3 x 2 x 1 = (k+1) x k!      This is a key observation!
(k+1)! = (k+1) x k!
(k+1) x k! > (k+1) x k²       By our Induction Hypothesis
(k+1) x k² = k³ + k² 
k³+k² > k² + 2k + 1    Because k³ > 2k+1 for k>4
k³+k²> (k+1)² 

Now linking up the chain of inequalities from beginning to end yields (k+1)! > (k+1)² Which is the original inequality with n=k+1.
Hence our inductive step holds. Practically this means that if we know the inequality holds for n=k, then it also holds for n=k+1

The final step is to note that since our base case (n=4) holds. And our inductive step (n=k+1) holds, it follows that n=5 holds. Since n=5 holds, and n=k+1 holds, sp does n = 6, we can repeat this indefinitely, showing that any n greater than or equal to 4 satisfies the given inequality.