nt.number theory - Is a "non-analytic" proof of Dirichlet's theorem on primes known or possible?

I was struggling to find an elementary proof of Dirichlet's theorem using another interesting technique.
I came finally to a proof from an entirely different direction but as i found out Erdos came first before many years.
The proof is not well known (I never understood why anyone mentions it!)and it uses Chebyshev type estimates. Here is the proof: http://kam.mff.cuni.cz/~klazar/ln_antcII.pdf
I hope this will be helpfull!

We are looking for prime numbers of the form $a+nm$.$(a,m)=1,n=1,2,...$

The general plan is:
$n!$ divides the product of $n$ consecutive terms of the arithmetic proggression $a+m$,$a+2m$,$...,$$a+nm$ with $(a,m)=1$ (if we disregard the factor of $n!$ which includes divisors of $m$)

For example, consider the proggression $1+3m$:

$4cdot7cdot10cdot13cdot16cdot19cdot22$ is divided by $7!/3^2$
(The factor that we "disregard" is $3^2$)

It is easy to prove in analog with Legendre's Lemma at binomial coefficients, that the highest power of a prime $p$ which divides $frac{(a+m)cdot(a+2m)...cdot(a+nm)}{n!}$ does not exceed $a+nm$.

The problem is that such a prime $p$ could not be of the form that is wanted.For example turning back to $frac{4cdot7cdot10cdot13cdot16cdot19cdot22}{7!/3^2}$ we find 11 as a divisor which is a prime of the form $2+3m$.

We wish to simplify the unwanted primes from the "big" fraction which Erdos calls $Pn(a,m)$.
In order to do this we divide $Pn(a,m)$ with a fraction of the same kind but from another proggression $a'+nm$ with $(a',m)=(a,m)=1$.

(The "other" proggression for $1+3m$ is $2+3m$ since only 1 and 2 are the only numbers coprime to 3 and less than 3).

But we find that in $Pn(a,m)$ every prime of the form $a'+km$ which is greater than $n$ exists exactly once, so (and here is the big idea) dividing $Pn(a,m)$ with $P(n/h)(a',m)$ ($h$ is the number with the property $a'h=a(modm)$ ) every prime of the form $a'+km$ which is greater than $n$ cancels.

Continuing like this you can cancel every unusefull prime that exceeds n and have only "small" unusefull primes whose product is significally smaller than $Pn(a,m)$.

With this,you prove that primes of the form $a+nm$ have a product which tends to infinity and so they are infinite.

I hope this is helpfull.

(note)We use $h$ because the smallest term of the proggression $a+nm$ that is divided by a prime $p$ of the form $a'+km$ is the term $a+km=hcdot p$ .