<s>[INST] Where is the flaw in my proof? Possible proof of the conjecture that there always exists a prime between $n^2$ and $(n+1)^2$?

Assume the statement is invalid. Therefore, for some $n \in \mathbb{N}$, there does not exist a prime that ranges from $n^2$ and $(n+1)^2$ exclusive. Let $\pi (x)$ denote the amount of prime numbers less than or equal to $x$, then we have that if $\pi ((n + 1)^2) - \pi (n^2) = k$, there exists $k$ primes that range between $n^2$ and $(n + 1)^2$ exclusive. Therefore, by assuming the conjecture is false, we consider $k = 0$, which consequently gives us the following: $$\begin{align} \pi ((n + 1)^2) - \pi (n^2) &= 0 \\ \Leftrightarrow \pi ((n+1)^2) &= \pi (n^2). \end{align}$$ It is well known that $\pi (x) \approx \dfrac{x}{\ln x}$ so by order of substitution, $$\frac{n^2}{\ln n^2} \approx \frac{(n + 1)^2}{\ln (n + 1)^2}.$$ For some $m$, one can see that $\ln x^2 = m \Rightarrow e^m = x^2 \Rightarrow \sqrt {e^m} = e^{m/2} = x \Rightarrow \ln x = m/2$. Therefore, we conclude that $\ln x^2 = 2\ln x$. $$\begin{align} \therefore \frac{n^2}{2\ln n} &\approx \frac{(n + 1)^2}{2\ln (n + 1)} \\ \\ \Leftrightarrow \bigg(\frac n2\bigg)\bigg(\frac{n}{\ln n}\bigg) &\approx \bigg(\frac{n + 1}{2}\bigg)\bigg(\frac{n + 1}{\ln (n + 1)}\bigg) \\ \\ \Leftrightarrow \bigg(\frac n2\bigg)\pi (n) &\approx \bigg(\frac{n + 1}{2}\bigg)\pi (n + 1) \\ \\ \Leftrightarrow \frac{n}{n + 1} &\approx \frac{\pi (n + 1)}{\pi (n)}. \end{align}$$
Now, by taking the reciprocal of the equation, it follows then that, $$\frac{n + 1}{n } = 1 + \frac 1n \approx \frac{\pi (n)}{\pi (n + 1)}.$$ However, although true, $$\square \ \dfrac{n + 1}{n} > \dfrac{\pi (n)}{\pi (n + 1)}.$$
By definition of the function $\pi$, we have that $\pi (n) = \pi (n + 1)$ if and only if $n$ and $n + 1$ are not prime and $\pi (n) < \pi (n + 1)$ if and only if $n + 1$ is prime, therefore $\pi (n) \leqslant \pi (n + 1)$. We thus arrive at the following conclusion: $$\begin{align} \frac{\pi (n)}{\pi (n + 1)} &\leqslant 1 \\ \\ \Leftrightarrow \frac{n + 1}{n} &> \frac{\pi (n)}{\pi (n + 1)} \\ \\ \Leftrightarrow \pi ((n + 1)^2) &> \pi (n^2) \\ \\ \Leftrightarrow \pi ((n + 1)^2) - \pi (n^2) &\geqslant 1 \\ \\ \therefore k &= 1. \end{align}$$ Thus, the conjecture is true.$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\bigcirc$

This looks correct to me, but this is a famous conjecture because it is apparently very difficult to prove and if true, it will tell us more about how the prime numbers are distributed across the positive integers. Therefore, I must have some kind of flaw in my proof, because it was too easy (only one page long). Surely this cannot be the case.
Thank you in advance.
 [/INST] You write $1+\frac{1}{n} \approx \frac{\pi(n)}{\pi(n+1)}$, which is equivalent to
$$
\lim_{n\to \infty}\frac{1+\frac{1}{n}}{\frac{\pi(n)}{\pi(n+1)}}=1,
$$
which is true because both numerator and denominator have limit $1$.
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