A series of questions. Explanations would be useful. I have done the first four parts. I am confused on how to go about the last two.
Show that for any prime $p$ the largest power of $p$ that divides $n!$ is $ \left\lfloor \frac{n}{p}\right\rfloor +\left\lfloor \frac{n}{p^2} \right\rfloor +\cdots +\left\lfloor\frac{n}{p^r}\right\rfloor$ where $p^r\le n < p^{r+1}$
Use the basic definition (no induction) to show that for any $m \ge 1$, $\lfloor \frac{2n}{m}\rfloor \le 2\lfloor\frac{n}{m}\rfloor+1$
Use the formula for $\binom{2n}{n}$ and the results of parts (1) and (2) to show that for any prime p, the largest power $p^r$ of $p$ that divides $\binom{2n}{n}$ satisfies $p^r\le 2n$
Prove that for any integer $n\ge 1$, $\binom{2n}{n}\ge 2^n$
Use the lower bound on the size of $\binom{2n}{n}$from part (4) and upper bound on each of its prime power factors from part (3) to prove that the number of distinct primes dividing $\binom{2n}{n}$ is at least $n/\log_2(2n)$
Conclude that there are at least $n/\log_2(2n)$ primes less than $2n$