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How to prove that there is no formula for $n!$ that does only use the binary operations $+,-,*,/$ on real numbers, powers of real numbers, and fixed real numbers?

Similar question have been asked several times on math.SE but none of the answers there provide a formula of the specified type. The same question over the naturals (where the result of division is a quotient) surprisingly has a positive solution.

I am quite sure that there is no such formula but I have no clue how one could prove that.

In particular, I do not allow integer parts (floors or ceilings), I do not allow integrals or derivatives, sum or product symbols, I do not allow substitutions into functions with multiple variables etc. The length of the formula should be uniformly bounded for all $n$ (otherwise $1\cdot2\cdot\ldots\cdot n$ would work).

My motivation is that my 9-year-old son asked me this question. This doesn't mean that your solution should be understandable by a 9-year-old, but rather that I only allow the operations that he has heard of. Also note that most likely there is no such formula but I want a proof for that.

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    $\begingroup$ Are we allowed to use rationals in exponential? for example $(n^{1/2})^{(1/2)^n}$? Also allowing arbitrary real numbers sounds to me like a bad idea - if there is a solution is likely to be with rather contrived real numbers and if there isn't it will be harder to prove. After all Mills' constant sounds amazing until you read how it is defined. @fe4's variant formalization with $k=\mathbb R$ captures this variant. $\endgroup$
    – i9Fn
    Commented 16 hours ago
  • $\begingroup$ Rationals are surely allowed in the exponent and I think contrived reals are welcome too, but of course I'm happy to see solutions to other variants as well. $\endgroup$
    – domotorp
    Commented 13 hours ago
  • $\begingroup$ Is something like $(-1)^{\frac12} = i$ forbidden? $\endgroup$
    – Daniel Weber
    Commented 3 hours ago
  • $\begingroup$ By default, yes, but I'm happy to see a solution to any variant of the problem. $\endgroup$
    – domotorp
    Commented 3 hours ago

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