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I was thinking about $\pi$ then thought: are there any formulas out there that require $\pi^n$ where $n \in \mathbb{N}$?

For example $\pi^2$ or $\pi^3$, but not just $\pi$? So are there any out there?

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    Every formula that includes $\pi$ includes $\pi^2$. **Proof:** $$\pi = \pi^{-1}\pi^{2}=\pi^{-1+2}$$. In general: $$\pi = \pi^{-n}\pi^{n+1}=\pi^{-n+n+1}=\pi^{1}$$ – 2017-02-13
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    Given the definition of the exponential function, $\mathrm e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$, the famous formula $\mathrm e^{\mathrm i\pi}+1=0$ contains all powers of $\pi$. – 2017-02-14
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    Are you interested in a power like $-1$? If so, look up Buffon's needle, a rather interesting result. https://en.wikipedia.org/wiki/Buffon's_needle – 2017-02-14

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Yes, though I'm assuming your asking after interesting formulas.

Some examples are $$\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$$ As seen from the Basel Problem. Similarly, $$\sum_{n=1}^{\infty} \frac{1}{n^4}=\frac{\pi^4}{90}$$ And so on. As mentioned by @imranfat, generally $$\sum_{k=1}^{\infty} \frac{1}{k^{2n}}=\zeta (2n)=(-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}$$ Where $\zeta(n)$ denotes the Riemann zeta function.

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    And in addition to that, other even powers of $\pi$ also arise in the answer when the $n$ receives higher even powers. – 2017-02-13
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    @imranfat Edited. – 2017-02-13
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    No, I am not asking after interesting ones, but the simple ones, but I really like the formulas you have provided within your answer. – 2017-02-15
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    @Simplestmathematics You're welcome. But if you are not looking for interesting ones then.. $$\pi^2= \pi \times \pi$$:) Glad to be of help though. – 2017-02-15
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I imagine that you have seen $\pi$ arise while calculating, for example, the area of a circle: $A=\pi r^2$. In fact, if you want to compute the volume of higher dimensional spheres you must calculate higher powers of $\pi$. For example the volume of an 8 dimensional sphere is given by

$$V=\frac{1}{24} \pi^4 r^8.$$

The Wiki page on n-sphere has a nice little section on this subject! In general we have that the volume of an $n$-dimensional sphere is proportional to

$$V(S_n) \propto \pi^{\lfloor n/2\rfloor},$$

so you have formulas involving $\pi$ to any power you desire, provided the dimension is sufficiently large (here $\lfloor \cdot \rfloor$ is the floor function).

  • $\frac{1}{2}{\pi}^2r^4\;\;$ volume of the $4$-dimensional sphere
  • $\frac{8}{15}{\pi}^2r^5\;\;$ volume of the $5$-dimensional sphere
  • $\frac{1}{6}{\pi}^3r^6\;\;$ volume of the $6$-dimensional sphere
  • $\frac{16}{105}{\pi}^3r^7\;\;$ volume of the $7$-dimensional sphere
  • $\frac{1}{24}{\pi}^4r^8\;\;$ volume of the $8$-dimensional sphere
  • $\frac{32}{945}{\pi}^4r^9\;\;$ volume of the $9$-dimensional sphere
  • $\frac{1}{120}{\pi}^5r^{10}\;\;$ volume of the $10$-dimensional sphere
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    Sure thing! Is there a way that I can allow you access to my answer? – 2017-02-13
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    There is no need to venture into higher dimensions. An ordinary torus has $\pi^2$ in its volume and surface area. – 2017-02-13
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    @IvanNeretin That is interesting to know. I wonder if one can (in a natural enough way) take higher genus surfaces to get higher powers of $\pi$? Or is that simply a pipe dream? – 2017-02-13
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The probability distribution function for the normal distribution has $\pi^{-\frac{1}{2}}$ in it: $$f(x) = \frac{1}{\sqrt{2\pi \sigma}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$

[I figured I'd answer this under 'community wiki' since there are going to be lots of suggestions.]

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    If you take multi-variate normal variable, the density will involve higher powers of $\pi.$ – 2017-02-14
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If $\Gamma$ denotes the Gamma function, the volume of the unit ball (open or close) in ${\mathbb{R}}^n$ is $$ m(B_n(0,1)) = \frac{{\pi}^{\frac{n}{2}}}{\Gamma(\frac{n}{2} + 1)}\mbox{,} $$ so, if $n = 2 k$, being $k \in \mathbb{N}$, $$ m(B_n(0,1)) = \frac{{\pi}^k}{\Gamma(k + 1)} = \frac{{\pi}^k}{k !}\mbox{.} $$

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Here is a famous (nevertheless surprising!) one for you: $$\int_{\mathbb R} e^{-x^2} dx = \pi^{1/2} $$ More generally: $$\int_{\mathbb R^n} e^{-|x|^2} dx = \pi^{n/2} $$

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    Corrected, thanks! And yes, that comment is not needed after all. I just remember being surprised at the simplicity of the generalization when first learning about this integral – 2017-02-14
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For a formula including $\pi^3$: $$ \int_0^{\frac{\pi}{2}}(\ln (\sin x))^2dx = \int_0^{\frac{\pi}{2}}(\ln (\cos x))^2dx = \frac{\pi}{2}(\ln 2)^2 +\frac{\pi^3}{24} $$

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The Stirling asymptotic formula for the factorial is

$$n!\approx\sqrt{2\pi n}\left(\frac ne\right)^n.$$

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You can find $\pi$ in BUCKLING:

https://en.wikipedia.org/wiki/Buckling