Is $b^{m+1} = O(b^m)$

Question

For $b$ and $m$ being variables, is it the case that $b^{m+1} = O(b^m)$ ? Any help would be much appreciated.

Answer 1

Remember that $$O(f(x))=g(x)\implies|f(x)|\le Mg(x)$$Where $M$ is a number and $x>x_0$ for some number $x_0$. Now, putting in $b^m$: $$O(b^m)=g(b)\implies|b^m|\le Mg(b)$$If we put $g(x)=b^{m+1}$, $M>0$, and $x_0=\dfrac{1}{M}$, then the condition holds.