Defying conventional calculus!

Which is the only positive integer whose \(k^\text{th}\) arithmetic derivative (where \(k \in \mathbb N\) ) is always equal to the number itself?

Details and Assumptions:

The arithmetic derivative function, denoted by \(n'\) is a function \(n: \mathbb{N} \cup \left\{0 \right\} \to \mathbb{N} \cup \left\{0 \right\}\) defined by

  • \(n'=1\) for all prime numbers.
  • \((ab)'=a' b+ a b'\) for \(a, b \in \mathbb{N} \cup \left\{0 \right\}\).
  • \( n\overbrace { '''....'' }^{ { k \ times } } = (((n\overbrace { ')')'....')' }^{ { k \ times } } \)
  • \(0'=1'=0\)

Clarification:

  • The set of positive integers (or counting numbers or natural numbers) do not contain the member \(0\) in them.
  • To be clear, it is asked to find a positive integer \(n\) which satisfies \[n = n' = n'' = n''' = \cdots = n\overbrace { '''....'' }^{ { k \ times } }\]

Inspiration.

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