Proof

Let $n > 1 $ be a natural number with the canonical representation. \[n=\prod_{i=1}^\infty p_i^{e_i},\quad (e_i\ge 0 )\] i.e. all $p_i$ are different and consecutive prime numbers with natural exponents $e_i\ge 0$.
Every other number \[d=\prod_{i=1}^\infty p_i^{\delta_i},\quad (\delta_i\ge 0 )\] is a divisor of $n$ if and only if $\delta_i\le e_i.$
Since there are $e_i+1$ possibilities for each exponent $\delta_i$ to be both, $\ge 0$ and $\le e_i$ and a specific choice of the exponent of one prime number is independent from the specific choice of the exponent of another prime number to find another divisor of $n$, it follows from the fundamental counting principle that the number of all distinct positive divisors must be $\tau(n)=\prod_{i=1}^\infty (e_i+1).$
The postulated formula for the number of of positive divisors follows.
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Scheid Harald: "Zahlentheorie", Spektrum Akademischer Verlag, 2003, 3rd Edition