Did You Know This ?????

If we say three consecutive whole numbers, that does not mean the first three. Similarly, nor should we assume its the first $r$ natural numbers. I proceeded by trying to find:

The product $\displaystyle P = a(a+1)(a+2)(a+3)...(a+r-1) = \prod_{i=0}^{r-1}(a+i)$ , where $a \in \mathbb{N}.$

Because if $a$ is even, $a + 1$ is odd, or if $a$ is odd, $a+1$ is even, we know that every other product is even. Thus, $P$ is divisble by $2$ raised to $r/2$ rounded down.

$2^{\lfloor r/2 \rfloor} \mid P.$

But $P$ also contains a product divisible by $4$ every 4th term, which must be counted an extra time, and a product divisible by $8$ every 8th term, which must be counted extra on top of that, and so on, for every power of $2$ .

Thus, $2^{\lfloor r/2 \rfloor}2^{\lfloor r/4 \rfloor}2^{\lfloor r/8 \rfloor}... \mid P$ , or in other words, $2^{\lfloor r/2 \rfloor+\lfloor r/4 \rfloor+\lfloor r/8 \rfloor+...} \mid P$ .

If not for the floor marks ( $\lfloor\rfloor$ ), we would be able to simply add up all of those fractional exponents, combining them into $2^r$ , but we cannot, because the floor function rounds down.

This is where I am stuck. We should have a solution a little less than $2^r$ , but I don't know how to proceed.