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Let $n=p_1^{a_1}{p}_2^{a_2}\cdots p_r^{a_r},a_i\ge 0$ be the prime factorization of the number $n$. Then the number of divisors of $n$ is $(a_1+1)(a_2+1)\cdots (a_r+1)$.

Let $k$ be a positive integer and let $f(2^k)=p_1^{a_1}{p}_2^{a_2}\cdots p_r^{a_r}$. Then $a_i=2^{b_i}-1$ for some integers $b_i\ge 0$, $i=1,2,\dots ,r$ and ${\sum }_{i=1}^r{b}_i=k$. Let $p_i,p_j$ be any two distinct prime factors of $f(2^k)$. If we replace $p_i^{2^{b_i}-1}$ with $p_i^{2^{b_i+1}-1}$ and $p_j^{2^{b_j}-1}$ with $p_j^{2^{b_j-1}-1}$ in the prime factorization of $f(2^k)$ we get a number that also has $2^k$ divisors (as $f(2^k)$) because $2^{b_i+1}\cdot 2^{b_j-1}=2^{b_i}\cdot 2^{b_j}$. Minimality of $f(2^k)$ implies $$p_i^{2^{b_i}-1}\cdot p_j^{2^{b_j}-1}<p_i^{2^{b_i+1}-1}\cdot p_j^{2^{b_j-1}-1},$$ or equivalently $$p_j^{2^{b_j}-1}<p_i^{2^{b_i}}.(\bullet )$$ If we allow some of the integers $c_i$ or $b_i$ to be zero, we can write $f(2^{k+1})=p_1^{2^{c_1}-1}{p}_2^{2^{c_2}-1}\cdots p_r^{2^{c_r}-1}$. Since ${\sum }_{i=1}^r{c}_i=k+1>k={\sum }_{i=1}^r{b}_i$, there is a positive integer $s$ such that $c_s>b_s$. Let $t$ be a positive integer different than $s$. Using $(\bullet )$ for the pair $(p_s,p_t)$, and for the pair $(p_t,p_s)$ gives $$p_t^{2^{c_t}}>p_s^{2^{c_s}-1}\ge p_s^{2^{b_s}}>p_t^{2^{b_t}-1}.$$ We conclude $c_t>b_t-1$, i.e. $c_t\ge b_t$ for all $t=1,\dots ,r$, which implies that $f(2^{k+1})$ divides $f(2^k)$.