On the number of prime numbers between $n^2$ and ${(n+1)}^2$

Let $p_{r+1}-1>n \geq p_r-1$, based on a sequence $\{1,2,3\cdots\ M_r(M_r=p_1p_2\cdots p_r)\}$, we compare the density of coprime numbers and establish a correlation between the proportions of coprime numbers in the ranges from 1 to consecutive square numbers. Then, we derive the relationship between the number of coprimes in the interval of $n^2 \sim {(n+1)}^2$ and the proportion of coprimes in the interval of $1 \sim n^2$, proving that there is at least one prime number between any $n^2$ and ${(n+1)}^2$. By extending our research to the range of $1 \sim M_r^2$, we establish the relationship between the proportions of backwards coprime numbers in the ranges from ${M_r}^2$ to consecutive square numbers; furthermore, we establish a relationship between the proportions of coprimes in small interval and the whole interval. Then, in conclusion, the number of coprimes between $n^2$ and ${(n+1)}^2$ is greater than $n\prod_{i=1}^{r}{(1-\frac{1}{p_i}})$, thus proving that there are at least 2 prime numbers between $n^2$ and ${(n+1)}^2$.