1. If $p$ is a prime and $k$ is a nonnegative integer, then $\tau(p^k)=k+1$
2. If $\gcd(a,b)=1$ then $\tau(ab)=\tau(a)\tau(b)$

Let $p$ be a prime. Then $p^0=1$ Note that 1 is the only positive divisor of 1 and $\tau(1)=\tau(p^0)=0+1=1$

Suppose that, for all positive integers $m$ smaller than $z \in \mathbb{Z}$ with $z>1$ the number of positive divisors of $m$ is $\tau(m)$ Since $z>1$ there exists a prime divisor $p$ of $z$ Let $k$ be a positive integer such that $p^k$ exactly divides $z$ Let $a$ be a positive integer such that $z=p^ka$ Then $\gcd(a,p)=1$ Thus, $\gcd(a,p^k)=1$ Since $1\le a<z$ by the induction hypothesis, there are $\tau(a)$ positive divisors of $a$

Let $d$ be a positive divisor of $z$ Let $y$ be a nonnegative integer such that $p^y$ exactly divides $d$ Thus, $0 \le y \le x$ and there are $k+1$ choices for $y$ Let $c$ be a positive integer such that $d=p^yc$ Then $\gcd(c,p)=1$ Since $c$ divides $d$ and $d$ divides $z$ we conclude that $c$ divides $z$ Since $c$ divides $p^ka$ and $\gcd(c,p)=1$ it must be the case that $c$ divides $a$ Thus, there are $\tau(a)$ choices for $c$ Since there are $k+1$ choices for $y$ and there are $\tau(a)$ choices for $c$ there are $(k+1)\tau(a)$ choices for $d$ Hence, there are $(k+1)\tau(a)$ positive divisors of $z$ Since $\tau(z)=\tau(p^ka)=\tau(p^k)\tau(a)=(k+1)\tau(a)$ it follows that, for every positive integer $n$ the number of positive divisors of $n$ is $\tau(n)$