The following is a picture from 'History of the theory of numbers, volume l Divisibility and Primality' by L. E. Dickson. Dickson wrote J. perott's proof of 'Infinitude of primes'. The first line does not seem correct to me.If $n=2$ and $N=3$ leads to contradiction. It seems Dickson did some mistake. Is not it?
Any help would be appreciated. Thanks in advance.
You are correct.
If $p_1, ...., p_n$ are the primes less than $N$ then the square free numbers that can be made with these prime factors can be formed by taking a product that can be made by either including a $p_k$ as a factor ... or not including it. So there are $2^n$ square free numbers that can be made.
(Example. For $p_1,p_2, p_3$ the $2^3 =8$ square free numbers that can be formed are $1, p_1=2; p_2= 3; p_3 = 5; p_1p_2 = 6; p_1p_3=10; p_2p_3 =15; p_1p_2p_3=30$.)
But many, maybe even most will be Larger than $N$. If $p_3 = 5 \le N < p_4 =7$ then three of those $8$ squarefree numbers are more than $N$
Indeed by Bertrands Postulate (which is hard to prove and maybe even unproven in Dickson's day) If $p_n \le N < p_{n+1}$ then $2p_n > N$.
So the statement that the squarefree numbers less than or equal to $N$ being $2^n$ is certainly false. But the number of square free numbers being less than $2^n$ is certainly true.
And that is enough for the rest of the proof.