$(\omega+1)\cdot \omega=\omega ^2$ by definition

Question

I'm trying to show that:

$(\omega+1)\cdot \omega=\omega ^2$ by definition:

$(\omega+1)\cdot \omega= \text{sup}\{(\omega+1)\cdot n: n \in \omega\}$ I can see is that $(\omega+1) \cdot n = (\omega+1)+(\omega+1)+...+(\omega+1)= \omega+\omega+...+\omega+1$

how do i prove that sup is $\omega^2$?

Answer 1

Hint: $$\omega\cdot n<(\omega+1)\cdot n<\omega\cdot (n+1)$$