Number of Primes is Infinite/Proof 6

Proof
Let $p_1, p_2, \ldots, p_j$ be the first $j$ primes.

For each real $x$ and natural number $j$, let:
 * $\map {M_j} x = \set {n \in \N \mid n \le x, \, n \text { is not divisible by any prime } p \text { with } p > p_j}$

Define:
 * $\map {N_j} x = \# \map {M_j} x$

Let $n \in \map {M_j} x$ for some $x$, $j$.

We can write:
 * $n = n_*^2 m$

where $n_*$ and $m$ are integers with $m$ square-free.

By the Fundamental Theorem of Arithmetic, we may uniquely write:
 * $\ds m = \prod_{i \mathop = 1}^j p_j^{e_j}$

where $e_j \in \set {0, 1}$.

Since there are $2$ choices for each $e_j$, there are $2^j$ possible values of $m$.

Note that since $m \ge 1$, we have:
 * $n_*^2 \le n \le x$

so:
 * $n_* \le \sqrt n \le \sqrt x$

So, there at most $\sqrt x$ possible values of $n_*$.

So, there are at most $2^j \sqrt x$ possible values of $n$.

That is:
 * $\map {N_j} x \le 2^j \sqrt x$

that there are only finitely many primes.

In particular, assume that there are $k$ primes.

Then:
 * $\map {N_k} x = x$

for each $x$, since there are no primes exceeding $p_k$.

So, we have:
 * $x \le 2^k \sqrt x$

for all $x$.

But this does not hold for all $x$, for example taking $x = 2^{2 k} + 1$.

So, we have arrived at a contradiction.

So there are infinitely many primes.