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It is considered standard in mathematics that all integers can be expressed as the product of primes: $$n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$$ Where $p_i$ is prime and $p_{i-1}<p_i$ and $a_i$ is and integer such that $a_i \geq 1$. But this means this does not explicitly define all integers in terms of primes, other integers must be used to describe the powers. Of course those integers can be expressed by a string of primes with powers and then those powers expressed by a string of primes with powers ... etc. So whilst the powers are really just multiplying primes by one another a certain number of times, has any work gone into looking at expressing numbers in multiplicative power form with just primes or $1$? In other words, can you keep substituting in prime chains in the place of the powers until all integers involved are primes or $1$?

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Yes, of course your assumption is correct.

Consider for example, $\large{n=3^{56}\cdot 5^{48}}.$

Let us colour exponents that are composite, and indent for each level of tetration for which the action must be repeated:

\begin{align} &\quad n&=&\quad 3^{\color\red {56}}\cdot 5^{\color\red {48}}\\ \text{tetration level 1:}&\quad \color\red {56}&=&\quad 2^3\cdot 7\\ &\quad \color\red {48}&=&\quad 3\cdot 2^{\color\red {4}}\\ \text{tetration level 2:}&\quad \color\red {4}&=&\quad 2^2\\ \Rightarrow&\quad \large{n}&=&\quad \large{3^{2^{3}\cdot 7}\cdot 5^{3\cdot 2^{2^2}}}\\ \end{align}

If it were not true (ie, if there were some stage at some level of tetration that factorisation yielded neither prime nor composite), it would of course contradict the fundamental theorem of arithmetic.

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    $\begingroup$ Thanks! I'll wait a little while and then mark this as accepted. $\endgroup$ – J-S Sep 8 '15 at 21:32
  • $\begingroup$ @J-S absolutely - that is usual. It gives others a chance to provide a better answer for a reasonable period of time :) $\endgroup$ – martin Sep 8 '15 at 22:53
  • $\begingroup$ @J-S BTW I don't know whether you use Mathematica, but on the back of this question, I asked a computational approach to the problem at MMA SE and yohbs wrote a nice little fuction here that outputs for most $n$ (unless it is too large - and it needs top be really quite big to be too big!! ;) $\endgroup$ – martin Sep 9 '15 at 0:36
  • $\begingroup$ I've never used Mathematica before but I'll have a look. Maybe convert it to python for a bit of fun :) $\endgroup$ – J-S Sep 9 '15 at 1:23
  • $\begingroup$ @J-S Would be interested to see that :) $\endgroup$ – martin Sep 9 '15 at 1:23

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