This is my last attempt to the Legendre's Conjecture: based on my first one, it's not that difficult to follow, I'm not using logical manipulation or something like this, it's all about inequalities and functions already proved! Thanks to all the readers!
Legendre's conjecture states that there is a prime number between $n^2$ and $(n + 1)^2$ for every positive integer $n$;
be $p$ the largest prime that divides $\frac{(n^2+2n)!}{(n^2)!}$;
be $p_i$ the i-th prime number;
According to Legendre's conjecture, there must be $n^2<p<(n+1)^2,∀ n \in \Bbb N$;
Let's take the product of every integer between $n^2$ and $(n+1)^2$, that will be $\frac{(n^2+2n)!}{(n^2)!}$, so $p$ it's a prime number that divides it;
Proof by contradiction
Let's suppose that $p$ is not between $n^2$ and $(n+1)^2$, so it must be $2≤p≤\frac{n^2+2n}{2}$
(In fact, if $\frac{n^2+2n}{2}<p<n^2$ it could not exist a $p_i$ such that $p*p_i<(n+1)^2)$;
As a consequence, every factor that divides $\frac{(n^2+2n)!}{(n^2)!}$ must be $2≤p_i≤p$ (because $p$ is the biggest!);
Now, how many prime numbers $p_i≤p$ MAY form $\frac{(n^2+2n)!}{(n^2)!}$?
I obviously don't know: however, I know how many prime numbers are between $2$ and $\frac{n^2+2n}{2}$:
$\pi(\frac{n^2+2n}{2})$, where $\pi(x)$ is the Prime-counting function;
Now, how I know how many times each prime form $\frac{(n^2+2n)!}{(n^2)!}$?
We will use
(1) $$\sum_{k=1}^\infty \left\lfloor\frac{n^2+2n}{p_i^k}\right\rfloor-\sum_{k=1}^\infty \left\lfloor\frac{n^2}{p_i^k}\right\rfloor=\sum_{k=1}^\infty \left\lfloor\frac{n^2+2n}{p_i^k}\right\rfloor-\left\lfloor\frac{n^2}{p_i^k}\right\rfloor$$
...and I will show show you that:
(2) $$\prod_{i=1}^A p_i^B<\frac{(n^2+2n)!}{(n^2)!}$$
$A=\pi(\frac{n^2+2n}{2})$
$B=\sum_{k=1}^\infty \left\lfloor\frac{n^2+2n}{p_i^k}\right\rfloor-\left\lfloor\frac{n^2}{p_i^k}\right\rfloor$
As I said, I don't know the prime numbers that form $\frac{(n^2+2n)!}{(n^2)!}$, so I will multiply EVERY single prime number at his highest possible exponent for that factorial, showing that even if I do this, the product you get is $<\frac{(n^2+2n)!}{(n^2)!}$, so you will need more factors to reach it, and because I have already multiplied every primes $≤\frac{n^2+2n}{2}$ at the highest exponent, the other factor(s) must be $>\frac{n^2+2n}{2}$, and because there are not factor $\frac{n^2+2n}{2}<p_i<n^2$, $p$ must be between $n^2$ and $(n+1)^2$;
The problems now are $\pi(\frac{n^2+2n}{2})$, $p_i$ and $B$:
For the first, Pierre Dusart proved that
(3) $$\pi(x)<\frac{x}{ln(x)-1.1}, x>60183$$
so, I will put
(4) $$\pi(\frac{n^2+2n}{2})=\frac{\frac{n^2+2n}{2}}{ln(\frac{n^2+2n}{2})-1.1},n>345$$
However, using this I will generate more primes than there really are, but even with this, the demonstration holds;
For the second, Rosser-Havil proved that
(5) $$i(ln(i \times ln(i)) - 1)<p_i<i \times ln(i \times ln(i))$$
The left holds for $i>0$, the right for $i>5$:
so, I will put
$$p_i=i(ln(i \times ln(i)) - 1)$$
with $i>3$, because for $1,2,3$ it gives negative results.
(Now you're thinking: why did you not put $p_i=i \times ln(i \times ln(i))$? That's because in that way I will generate a number that is less than the one I will generate using $p_i$, and it will mess up everythink, because I need to show that even moltiply highest value, the disequality holds);
For the third, is quite trivial that
(6) $$\sum_{k=1}^\infty \left\lfloor\frac{n^2+2n}{p_i^k}\right\rfloor-\left\lfloor\frac{n^2}{p_i^k}\right\rfloor≤\frac{2n}{p_i-1}$$
So, I will put
(7) $$\sum_{k=1}^\infty \left\lfloor\frac{n^2+2n}{p_i^k}\right\rfloor-\left\lfloor\frac{n^2}{p_i^k}\right\rfloor=\frac{2n}{p_i-1}$$
Let's return to (2): so it becomes:
(8) $$2^{2n} \times 3^{n} \times 5^{n/2} \times \prod_{i=4}^A C^B<\frac{(n^2+2n)!}{(n^2)!}$$
$A=\frac{\frac{n^2+2n}{2}}{ln(\frac{n^2+2n}{2})-1.1}$
$B=\frac{2n}{C-1}$
$C=i(ln(i \times ln(i)) - 1)$
(notice that $2,3,5$ are out of the multiplication, because they give negative results if I use $i(ln(i \times ln(i)) - 1)$);
Let's raise both sides to $1/{2n}$:
(9) $$2 \times 3^{1/2} \times 5^{1/4} \times \prod_{i=4}^A C^{\frac{1}{C-1}}<\left(\frac{(n^2+2n)!}{(n^2)!}\right)^{1/2n}$$
that becomes
(10) $$\frac{\left(\frac{(n^2+2n)!}{(n^2)!}\right)^{1/2n}}{2 \times 3^{1/2} \times 5^{1/4} \times \prod_{i=4}^A C^{\frac{1}{C-1}}}>1$$
and finally, we have to prove that
(11) $$\lim_{n\to +∞} \frac{\left(\frac{(n^2+2n)!}{(n^2)!}\right)^{1/2n}}{2 \times 3^{1/2} \times 5^{1/4} \times \prod_{i=4}^A C^{\frac{1}{C-1}}}>1$$
(We are not working on prime numbers anymore!)
This limit exists, but I'm not able to solve it: Mathematica says it's slightly bigger than $\sqrt{2}$, and because $\sqrt{2}>1$ We have done (Remember it holds for $n>345$);
So, as I said, I need more factor to reach $\frac{(n^2+2n)!}{(n^2)!}$ and we have a contradiction, because we supposed $2≤p≤\frac{n^2+2n}{2}$ so $p$ is bigger than expected, and it must be between $n^2$ and $(n+1)^2$.
You bound the number of 2s in the factorization of $\frac{(n^2+2n)!}{(n^2)!}$ by $2n$, but $$ \frac{(22^2+2\cdot22)!}{(22^2)!}=2^{47}3^{24}5^{12}\cdots $$ is divisible by $47>2\cdot22$ 2s.
Your bound at 3 fails at 7, your bound at 5 fails at 3, your bound at 7 fails at 2, and your bound at 101 fails at 10. Because your bounds aren't correct they can't be used to prove the desired result.
The main problem is the large primes, for which your bound says that there can't even be a single one. For example, with $n=1000$ and $p=1999$ (which is much smaller than $\frac{1000^2+2\cdot1000}{2}$), the bound is around 48 which is much less than 1999 and so you are effectively excluding that prime from appearing. But it does appear since 1001499 = 501*1999 and 1000000 < 1001499 < 1002000.