# Is it possible to show that $a\cdot a^{\lfloor\frac{b}{2}\rfloor}\cdot a^{\lfloor\frac{b}{2}\rfloor} = a^b$ when $b$ is odd

I have $$a$$ and $$b$$ and $$b$$ is odd $$a$$ is an integer and $$b$$ is a strictly positive integer. Is there a way I can show:

$$a\cdot a^{\lfloor\frac{b}{2}\rfloor}\cdot a^{\lfloor\frac{b}{2}\rfloor} = a^b$$

I know I can simplify to:

$$a^{2\lfloor\frac{b}{2}\rfloor + 1}$$ and that $$2\lfloor\frac{b}{2}\rfloor + 1$$ is in the form of an odd number $$2k+1$$. Is there any way I can simplify more?

If $$b$$ is odd, $$\lfloor \frac{b}{2}\rfloor = \frac{b-1}2.$$ Substitute that and you can simplify the expression.