Finding an $a$, such that $\forall x(x^x-a\cdot x!=0).$

Question

I've been finding myself wondering about this equation for a long time, however due to my limited math knowledge, I can't solve or even determine if there is a solution to that equation.

So I ask: is there an equation or number that can satisfy that?

Answer 1

from https://en.wikipedia.org/wiki/Stirling%27s_approximation#Speed_of_convergence_and_error_estimates we get an approximation that gets pretty good as $x$ gets large, $$ \frac{x^x}{x!} \approx \frac{e^x}{\sqrt{2 \pi x} \left( 1 + \frac{1}{12 x} + \frac{1}{288 x^2} - \frac{139}{51840 x^3} - \frac{571}{2488320 x^4} \right)} $$

Quite good, even with small values:

x   x^x                          x!      x^x / x!
1   1                             1       1                   approx 1.00050078212357
2   4                             2       2                   approx 2.000042039543573
3   27                            6       4.5                 approx 4.500013477631439
4   255.9999999999999            24      10.66666666666666   approx 10.66667448781807
5   3124.999999999999           120      26.04166666666666   approx 26.0416730195439
6   46656.00000000003           720      64.80000000000004   approx 64.80000640706673
7   823542.9999999994          5040     163.4013888888888   approx 163.401396402745
8   16777215.99999998         40320     416.1015873015867   approx 416.1015971490262
9   387420489.0000001        362880    1067.627008928572   approx 1067.62702298249
10   10000000000            3628800    2755.73192239859    approx 2755.731943854943
11   285311670610.9999     39916800    7147.658895778216   approx 7147.658930376819
12   8916100448255.988    479001600   18613.92623376621   approx 18613.92629213758
13   302875106592253.4   6227020800   48638.84613847017   approx 48638.84624076476

Answer 2

$\mathbf{Proposition}$

$\not \exists a\in \Bbb R\forall x(x^x=a\cdot x!$).

$\mathbf{Proof:}$

We are looking for an $a$ that works for all $x$'s. Let's look at two different particular cases: $x=2$, $x=1$.

Case 1: $x=2\implies x^x=4; a\cdot x!=2a \implies a=2$.

Case 2: $x=1\implies x^x=1; a\cdot x!=a\implies a=1$. As these two particular cases require different values of $a$, an $a$ that satisfies all the cases does not exist.