What's the closest approximation to $\pi$ using the digits $0-9$ only once? 
What's the closest approximation to $\pi$ achievable using each digit $0-9$ no more than once, and basic operations of roots, brackets, exponentiation, addition subtraction, concatenation, division and factorial?

This was mentioned in another question and I thought it was fun.  I can pretty quickly come up with (with a bit of help from Ramanujan):
$$\frac{7}{3}\left(1+\sqrt{\frac{6}{50}}\right)-\frac{8}{2\times4\times9!}\approx3.1415958$$
I challenge anybody to get beyond 20 decimal digits of accuracy!
 A: $$\frac{354 + (0!)^8}{\sqrt{12769}} = \frac{355}{113} \approx 3.1415929203$$ which is close to $\pi$ with an error absolute value of $2.6 \times 10^{-7}$.
This comes from one of the convergent approximations of the continued fraction expansion of $\pi$.
I am trying to find a slightly more accurate variation of this.
EDIT: This is a very liberal interpretation of the rules, but if we allow factorials of non-integers then we can obtain an exact formula for $\pi$, such as:
$$\pi = \left(\frac{3}{2}\right)!\times\left(\frac{6}{4}\right)!\times\frac{8}{9}\times(7-5)\times(1+0)$$
This uses the fact that $\left(\frac{3}{2}\right)! = \Gamma\left(\frac{5}{2}\right) = \frac{3\sqrt{\pi}}{4}$ (there's probably other nicer variations).
A: A simple exact expression using the gamma function: $$\left(\frac63\cdot\left(\frac48\right)!\right)^2$$
My answer below, based on tehtmi's answer, has two more decimal digits of accuracy. It applies a factor to the sum of the first five terms instead of adding two more terms.
$$(\sqrt{^5 70} - \sqrt{^{15} 9!} + \sqrt{^{23} 8!} + \sqrt{^{34} 14!} + \sqrt{^{36} .5}) * \sqrt{^{44} 6} * \sqrt{^{50} 2} * \sqrt{^{58} 3} \approx \pi - 1.926.. \times10^{-18} $$
The equivalent expression here can be plugged into Wolfram Alpha: (70^(.5^5)-9!^(.5^15)+8!^(.5^23)+14!^(.5^34)+.5^(.5^36))*6^(.5^44)*2^(.5^50)*3^(.5^58)
A: Taking inspiration from @Dan Brumleve's answer, we could attempt to use nested square-roots, factorials and concatenations (i.e. floor and ceiling functions) to find integers whose ratio is arbitrarily close to $\pi$.
Various people have previously studied whether you can get every integer from nests of roots, factorials and concatenations, possibly only using the number $3$. Here are a few links to discussions on the topic:


*

*Using floor, ceiling, square root, and factorial functions to get integers

*Approximating $\pi$ by an expression of the form $\sqrt{\sqrt{ \cdots \sqrt{ n!! \cdots !}}}$

*http://mathforum.org/wagon/current_solutions/s1171.html

*http://people.missouristate.edu/lesreid/pow11_02.html
I will be taking the floor of each square-root value, to make sure no factorials of non-integer arguments are taken. You don't have to do this, and you may find more numbers are possible when taking repeated square-roots. But you may need to be wary of only taking a factorial after a floor or ceiling function.
Define the string $3FSS\ldots$ to be the order of factorials ($F$) and floors of square-roots ($S$), read left to right. For example, $3FSFF$ would represent $\left(\left\lfloor\sqrt{3!}\right\rfloor!\right)!=2$. Also, for neatness' sake, let $S_n=\underbrace{S\ldots S}_n$
The following table shows some possible values with this method.
$$\begin{array}{|c|c|}
\hline
\text{number} & \text{generating string} & \text{algebra} \\ \hline
1 & 3S & \lfloor\sqrt{3}\rfloor \\
2 & 3FS & \lfloor\sqrt{3!}\rfloor \\
3 & 3 & 3 \\
4 & 3FFSFS_4FS_5FS_7 & \ldots \\
5 & 3FFSS & \left\lfloor\sqrt{\lfloor\sqrt{(3!)!}\rfloor}\right\rfloor \\
6 & 3F & \lfloor\sqrt{3}\rfloor \\
\hline
\end{array}
$$
Hence, we can say that: $$\pi\approx\frac{3F_2S_2FSFS_2FS_4}{3F_2SFS_4FS_5FS_6FS_3}=\frac{1989}{633}=\color{red}{3.14}218\ldots$$
Therefore, we can get the first $3$ digits using only the number $3$. This could easily be extended to avoiding using $3$ twice, by replacing $3F$ with $6$.
We could further extend this technique by:


*

*Nesting operations on the digits $4$ to $9$ too and combining the numbers in order to get a more efficient representation of $\pi$.

*Combining the nests of operations with different digits (e.g. $\sqrt{3!+4!}$).

*Proving whether all integers are reachable with this method, which would trivially prove that arbitrarily accurate approximations of $\pi$ are possible.


My very messy and un-pythonic python code used to generate the above formulae is provided here.
A: With 8 digits and with paper-and-pencil and Wolfram Alpha, I obtained:
$$\sqrt[8]{{7!}}+\frac {2}{9}+ \frac{5-4}{60} \approx 3.1416000$$
A: $\frac{\log (5280^{\sqrt{9}} + 3!! + 4! )}{1 \times \sqrt{67}} \approx 3.14159265358979324$, good for 18 places
If you don't like logs,
$(8\times9 +\frac{52-0!}{73})(\sqrt{-1}^{\sqrt{-4}}) \approx 3.14159266$, good for 8 places of accuracy.  
$\sqrt{\sqrt{\frac{2143}{5!!+7}}} \approx 3.141592653$, good for 9 places of accuracy.
$3 \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{20-\frac{4\times8}{9\times5 - 7 - 1}}}}}}} \approx 3.141592652$, good for 9 places of accuracy. 
For $e$, I challenge people to beat Sabey's approximation of 
$(1+9^{-4^{7\times6}})^{3^{2^{85}}}$, which is only accurate to 18457734525360901453873570 decimal digits.
A: I can't prove this, but I think we may be able to get as close as we want by starting with $1234567890$, taking some number of factorials, and then taking square roots until we get a number less than $\pi^2$.  Heuristically we get a random number in the interval $[\pi, \pi^2]$ whose logarithm is uniformly distributed so we ought to be able to get as close as we like to $\pi$ by taking enough factorials.
A: My answer uses lots of nested square roots, so I'll write $n$ nested square roots on $x$ as $\sqrt{^n x}$ to avoid writing them all out. Evaluated with Wolfram Alpha.
$$\sqrt{^5 70} - \sqrt{^{15} 9!} + \sqrt{^{23} 8!} + \sqrt{^{34} 14!} + \sqrt{^{36} .5} + \sqrt{^{41} 2} - \sqrt{^{46} .6} \approx \pi + 5.477.. \times10^{-16} $$
Here's something that should be equivalent that can be plugged into Wolfram Alpha: 70^(1/2^5) - 9!^(1/2^15) + 8!^(1/2^23) + 14!^(1/2^34) + (5/10)^(1/2^36) + 2^(1/2^41) - (6/10)^(1/2^46)
Basically, the fractional part of $\sqrt{^5 70}$ is a good approximation for the fractional part of $\pi$. Then the fractional part of $\sqrt{^{15} 9!}$ is a good approximation for the remaining fractional part error. And so on, with a bit of finesse to make the integer part work out nicely. Chaining a bunch of square roots on anything will always give something that helps at least a little, since you can get numbers arbitrarily close to $1$. I did a fairly simple greedy search.
Surely it is possible to do better with this approach, but I was about to the point where double precision floating point runs into trouble representing $(1 \pm \text{error})$ anyway.
A: Cheating a little with decimal points, but very, very simple:
$3.8415926 - 0.7$
Also:
$3 + 1/7 - (6/((9480/2)+5)) \approx 3.1415926539214 \approx \pi + 3.316 \times 10^{-10} $
A: Very "bad" approximation with all digits and only three elementary operations:
$$3+\frac{1}{7}+\frac{2 \cdot 5}{809 \cdot (6+4)} \approx 3.1440$$
