Proof
(related to Proposition: Calculating the Number of Positive Divisors)
 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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References
Bibliography
 Scheid Harald: "Zahlentheorie", Spektrum Akademischer Verlag, 2003, 3rd Edition