Prove every nonzero integer has only finitely many divisors 
Prove every nonzero integer has only finitely many divisors
axiom: For $d,n \in \mathbb{Z}$, if $d \mid n$, there  $\exists k \in \mathbb{Z}$ such that $d \cdot k = n$

$\forall d > n, \,\, k = \frac nd = \notin \mathbb{Z}$
$\forall d < -n, \,\, k = \frac n{-d} = \notin \mathbb{Z}$
$\longrightarrow d$
is bounded $ \longrightarrow \mid d \mid < \infty$
 A: Let $p_n$ be the prime factors and $q_n$ be the powers of those prime numbers.
We can break a number $r$, into the factored form $p_1^{q_1}p_2^{q^2}p_3^{q_3}$...
And yes, this series has infinite factors.
For example, we have $36=(2^2)(3^2)(5^0)(7^0)(11^0)....$
If we want to construct a divisor of $r$, we pick anywhere from $0$ to $q_1$ factors of $p_1$, we pick $0$ to $q_2$ factors of $p_2$, etc.
For $36$, we have the number of divisors is $(2+1)(2+1)(0+1)(0+1)(0+1)...=9$
Notice that if a number does not contain a particular prime factor, then the power of that prime factor is $0$ in the factorization, and only contributes a factor of $1$ to the product that represents the number of divisors.
Therefore, the number of divisors $r$ has is $\displaystyle (q_1+1)(q_2+1)(q_3+1)...=\prod_{k=1}^\infty(q_n+1)$.
We know that any finite number has a finite number of prime factors, each of which are raised to a finite power.
Therefore, we have that the product of a finite number of finite numbers multiplied by infinite $1s$ is also finite. √
