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Background: I was solving following question.

Which is bigger $2^{30!}$ or $\left(2^{30}\right)!$ ?

$2^{30!}$ can be expressed as $\left(2^{30}\right)^{29!}$

Note that for sufficiently large $n \in \mathbb{N}$ we have that $n! >> 2^n$.

Moreover, $\forall_{n \in \mathbb{N} \wedge n \ge 5}\ n! > 2^{n+1}$

Hence, $29! > 2^{30}$

Finally, $$2^{30!} = \left(2^{30}\right)^{29!} > \left(2^{30}\right)!$$

since $29! > 2^{30}$ and the fact every multiplicand of $\left(2^{30}\right)^{29!}$ is greater or equal than every multiplicand of $\left(2^{30}\right)!$.

(If you have other approaches please share them)

Then, I started thinking how would I approach something more generalised.

Question: For what $n \in \mathbb{N}$ and $r \in \mathbb{R}$ $n^r = n!$ ?

Some thoughts

Following the same argumentation implies that $\forall_{r > n}\ n^r > n!$.

This equation is trivially satisfied when $n = 1$ by all $r \in \mathbb{R}.$

Update

Courtesy of Shaun who pointed that $r = \log_n (n!)$ solves the equation $n^r = n!$.

  • 0
    Have you heard about [Stirling's approximation](https://en.wikipedia.org/wiki/Stirling%27s_approximation)? – 2017-02-02
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    So you suggest to substitute $n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ to approximate the solutions? – 2017-02-02
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    There's the case when $n=2, r=1$: $2^1=2=2\times 1=2!$. – 2017-02-02
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    You could make use of the [Gamma function](https://en.m.wikipedia.org/wiki/Gamma_function) for $n!$. – 2017-02-02
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    Consider when $r=\log_n(n!)$. – 2017-02-02
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    Thank you Shaun. I haven't spotted that $r = \log_n (n!)$ definitely solves the equation since the base of the power and base of the logarithm are the same. – 2017-02-02
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    Do you want an exact equality or only the approximate intersection point ? – 2017-02-02
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    @MaciejCaputa I misinterpreted you question. Please tell me if you find my answer useful nevertheless! – 2017-02-02

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For $n\ge 25$ , we have $$10^n

Hence

$$\large (2^{30})!<(2^{30})^{2^{30}}=2^{30\cdot 2^{30}}<2^{(10^{30})}<2^{30!}$$

Perhaps you can use the double inequality above for your generalized problem.

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    The magnitude of $(2^{30})!$ is roughly $10^{9\ 230\ 534\ 754}\approx10^{10^{10}}$, whereas $2^{30!}=10^{log_{10}(2)\cdot(30!)}\approx10^{8\cdot 10^{31}}$ – 2017-02-02
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    So, $\large 2^{30!}$ is far bigger – 2017-02-02
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    You can calculate those results here : https://mrob.com/pub/comp/hypercalc/hypercalc-javascript.html – 2017-02-02
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    I really like this thought process. The first equation you have might be applicable to lots of things. I will make sure to keep it in mind. – 2017-02-02
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    I could use the double inequality to estimate the values, however, the 3 functions diverge quite rapidly. – 2017-02-02
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    If the values don't get too big, you can still use the logarithm or even the double-logarithm. For very large numbers, we have $10^n\approx n^n$ in some sense (which seems to be absurd since $n^n$ is growing much faster than $10^n$). You can find this phenomen in the site above. So, if $n$ is large enough , we have $10^n\approx n!$. Use hypercalc to distinguish $(10^{10^{100}})!$ and $10^{10^{10^{100}}}$. You will see no difference! – 2017-02-02