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 .