Not a Calculus problem!

This isn't the most elegant solution but it creates a nice looking formula... $\displaystyle\prod_{r=1}^{n} (2r-1) = \frac{\displaystyle\prod_{r=1}^{2n} r}{\displaystyle\prod_{r=1}^{n} 2r } = \frac{(2n)!}{2^n (n!)} = \frac{n!}{2^n} {\ 2n \choose \ n }$ Now we can go further by showing that: $\ As \ n \rightarrow\infty \ , {\ 2n \choose n } \rightarrow\frac{4^n}{\sqrt{n \pi }}$ $\Rightarrow\lim_{n \rightarrow\infty} \displaystyle\prod_{r=1}^{n} (2r-1) = + \sqrt{\frac{n!(n-1)! 4^n}{\pi} }$ (Plugging in n=7 gives 135135 in the first formula and 137567.7... in the second)

Azhaghu Roopesh M posted this idea in the comments.

Note that this is the very definition of the odd double factorials. That is $\prod_{r=1}^n (2r-1) = (2n-1)!!$ We are given that the product is $135135.$ Note that $135135 = 13!!$ Then $(2n-1)!! = 13!! \\ 2n - 1 = 13 \\ \boxed{n = 7}$

Just Prime factorize it and bundle them to form Odd positive integers.

$135135=1 \times 3 \times 5 \times 7 \times 9 \times 11 \times 13$

Max odd integer=13

Hence $2n-1=13$ (at max $n=r$ )

$\therefore n=7$

As $135135=\dot { 100 } +35\equiv 10\left( mod\quad 25 \right)$ and $135135\equiv -1+3-5+1-3+5\left( mod\quad 11 \right) \equiv 0\left( mod\quad 11 \right)$ , it implies that $11\ge 2n-1>15\Rightarrow 6\ge n>8\Rightarrow n=6\quad or\quad 7$ . We can try three times, but $\prod _{ r=1 }^{ 6 }{ \left( 2r-1 \right) } =10395$ and $\prod _{ r=1 }^{ 7 }{ \left( 2r-1 \right) } =135135$

i think there are plenty of good solutions but i'm going to write the easiest : $\prod_{r=1}^{n}(2r-1)=((1*2)-1)((2*2)-1)((3*2)-1)((4*2)-1)........=1*3*5*7*.......$ and this pattern continues , i.e , multiplying out odd numbers to get 135135 Now , by multiplying the first few odd integers we got from the expansion ,and dividing 135135 by the answer we get $1*3*5*7=105 , \frac{135135}{105}=1287$ we next divide 1287 by the next odd integer , which is 9 $\frac{1287}{9}=143$ , we then divide this answer by the next odd integer , which is 11 $\frac{143}{11}=13$ and the number we get is the next odd integer in the progression , so that implies that the number 135135 could be written like $135135=1*3*5*7*9*11*13$ which would imply that $2r-1=13$ $2r=13+1=14$ that is $r=\frac{14}{2}=7$ which means that $n=7$ or in other words , that value of r=7 would give 13 .

$135135=1\times 3\times 5\times 7\times 9\times 11\times 13$

$2r-1=13$

$r=7$