Is there any way to find out $x! = n$ by hand? I've given myself this question but I can't see how to get it. I'm actually not taking calculus I'm in high school taking Algebra 2 but I've learned a bit or two from this site.

I did:

$x! = 10$

Trying to solve for $x$ but I had trouble doing so. Can anyone give me a hand here?
Wolfram alpha gives the roots of $\Gamma(x+1)=10$ as
$-4.995806,
-4.016334,
-2.947296,
-2.097191,
-1.095325,
3.390078.$

One root of $\Gamma(y)=x$ is approximated by $\frac{L(x)}{W(\frac{L(x)}e)}+\frac{1}2$ where $L(x)=\ln(\frac{x+c}{\sqrt{2\pi}})$, $c\approx0.036534$ and $W(x)$ is the principal branch of the inverse of $xe^x$. $W(x)$ can be approximated by selecting an intial $w_0$, and find succesive approximations $w_{j+1}=w_j-\frac{w_je^{w_j}-z}{e^{w_j}+w_je^{w_j}}$. $e^x$ and $\ln {1+x}$ can be found using $1+x+ \frac{x^2}/{2!}+\frac{x^3}{3!}...$ and $x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+...$ respectively. This can be done by hand but is tedious.
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EDIT: The expansion for $\ln(x+1)$ only works if $|x|&lt;1$. Otherwise use $\ln x\approx \frac{\pi}{2M(1,\frac{4}s}-m\ln 2$ with $M(1,\frac{4}s)=$ thearithmetic geometric mean of $1$ and $\frac{4}s$ ,$s=x2^m$ and $m$ any chosen integer (larger $m$ give a closer approximation).
If you allow for approximate solutions, in other words, you interpret "=" as meaning "approximately equal to", then you might consider 3 as a solution, since 3!=6, and 4!=24.