Why isn't $1$ a superior highly composite number?

Question

A superior highly composite number is a positive integer $n$ for which there is an $\epsilon>0$ such that $\dfrac{d(n)}{n^\epsilon} \geq \dfrac{d(k)}{k^\epsilon}$ for all $k>1$, where the function $d(n)$ counts the divisors of $n$.

The first superior highly composite numbers are $2, 6, 12, 60, 120, 360, 2520, 5040, 55440, \dots$

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Edit: Bump. I am still looking for answers.

Answer 1

The reason may be found in Ramanujan's original paper. He gives the following theorem:

If $N$ is a highly composite number, and $N = 2^{a_2} \times \dots \times p_1^{a_{p_1}}$ be its prime factorization, then $a_{p_1}=1$ except for $N=4$ and $N=36$.

This theorem fails if $N=1$ is considered to be highly composite (however theorem may not make sense if $N=1$). His definition of highly composite number is verbatim:

A number $N$ may be said to be a highly composite number, if $d(N') < d(N)$ for all values of $N’$ less than $N$.

With this definition $1$ should be highly composite, but $1$ does not appear in his list of highly composite numbers. So I suppose the author is implicitly considering numbers with at least one prime divisor (i.e. for which $p_1$ make sense in all his proofs). Same reason for why $1$ is not superior highly composite.