The divisor function \(\sigma_0(n)\) gives the number of divisors of a natural number \(n\).

Thus, \(\sigma_0(1)=1,\sigma_0(2)=2,\sigma_0(4)=3\)

Consider the function \(n=\sigma_{0,\min}^{-1}(k)\), where \(n\) is the smallest natural number with exactly \(k\) divisors.

Is the following statement true?

\[0<k_1<k_2 \implies \ \sigma_{0,\min}^{-1}(k_1) < \sigma_{0,\min}^{-1}(k_2)\]

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