# 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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### References

#### Bibliography

1. Scheid Harald: "Zahlentheorie", Spektrum Akademischer Verlag, 2003, 3rd Edition