Prove the following inequality regarding natural numbers

Question

For all natural numbers $a>2$. I want to prove, $$ (a-1)! + 1 < a^{a-1} $$ How do I go about this?

Answer 1

One method that you can use would be mathematical induction, which if you are not familiar with can be found here https://www.mathsisfun.com/algebra/mathematical-induction.html.

To start you off, let your base case be $a=3$ as that is the smallest natural number $>2$.

You can show easily $$(3-1)! + 1 < 3^{3-1}$$

Next you can let your equation $(a-1)! + 1 < a^{a-1}$ hold for all $a=k$, $a>2$,$a\in \mathbb{N}$ then you want to show that your equation still holds for all $a=k+1$. This will show you that for each $a$, the next $a$ will also hold, thus proving your equation.

I suggest visiting the linked site to learn a bit more about the method and then try attempting it yourself!

Good luck!

Answer 2

Simply note that $(a-1)!$ is the product of $a-1$ positive integers, each at most $a-1$. So $$(a-1)!+1\leq (a-1)^{a-1}+1\leq(a-1)a^{a-2}+1=a^{a-1}-a+1<a^{a-1}.$$