Number of Primes is Infinite/Proof 3

Proof
that there are only $N$ prime numbers.

Let the set of all primes be:
 * $\Bbb P = \set {p_1, p_2, \ldots, p_N}$

By the Fundamental Theorem of Arithmetic, every integer $k > 1$ can be expressed in the form:
 * $k = {p_1}^{a_1} {p_2}^{a_2} \dotsm {p_N}^{a_N}$

Let $n > 1$ be fixed.

Let $a$ be the largest exponent occurring in the prime decomposition of all positive integers $k \le n$.

Then:
 * $\displaystyle \sum_{k \mathop = 1}^n \frac 1 k \le \prod_{j \mathop = 1}^N \paren {\sum_{k \mathop = 0}^a \frac 1 { {p_j}^k} }$

is expressible in the form:

which can be seen by multiplying out the factors on the.

But from Sum of Infinite Geometric Progression:
 * $1 + x + x^2 + \dotsb = \dfrac 1 {1 - x}$

for all $x$ such that $\size x < 1$.

Thus the factors in $(1)$ are less than the numbers:
 * $\dfrac 1 {1 - 1 / p_1}, \dfrac 1 {1 - 1 / p_2}, \dotsb, \dfrac 1 {1 - 1 / p_N}$

and so:
 * $1 + \dfrac 1 2 + \dfrac 1 3 + \dotsb + \dfrac 1 n < \dfrac {p_1} {p_1 - 1} \dfrac {p_2} {p_2 - 1} \dotsm \dfrac {p_N} {p_N - 1}$

We have chosen $n > 1$ arbitrarily, so this holds for every $n > 1$.

This contradicts the result Harmonic Series is Divergent.

Hence the result, by Proof by Contradiction.