Wikipedia claims this but only cites an offline proof:
$$\lim_{n\to\infty} 2^n \sqrt{2-\sqrt{2+\cdots+ \sqrt 2}} = \pi$$
for $n$ square roots and one minus sign. The formula is not the "usual" one, like Taylor series or something like that, so I can't easily prove it. I wrote a little script to calculate it and it's clearly visible, but that's not a proof.
Best Answer
It's the sequence of approximations obtained when you approximate the perimeter of the circle of diameter $1$ with inscribed regular $n$-gons for $n$ a power of $2$.
As I happen to have this TeXed' up, I'll offer:
Suppose regular $2^n$-gons are inscribed in a circle of radius $r$.
Suppose the side length(the length of one "face") $a_n$ of a the inscribed $2^{n}$-gon is known (so, $a_2$ is the side length of the square). To find the side length of the $2^{n+1}$-gon, one may apply the Pythagorean Theorem twice to obtain $$ \tag{1}a_{n+1} = r\sqrt{2-\sqrt{4-{a_n^2\over r^2}}} $$
Now, starting with a square, $$a_2=\sqrt 2 r.$$ Using the recursion formula (1) repeatedly gives: $$ a_3%= r\sqrt{2-\sqrt{4-{2r^2\over r^2}}} = r\sqrt{2-\sqrt2}, $$ $$ a_4%= r\sqrt{2-\sqrt{4-{ ( r\sqrt{2-\sqrt2})^2 \over r^2}}} = r\sqrt{2-\sqrt{4-{ ({2-\sqrt2} ) }}} = r\sqrt{2-\sqrt{{ {2+\sqrt2} }}}, $$ and $$ a_5%= r\sqrt{2-\sqrt{4-{ ( r\sqrt{2-\sqrt{{ {2+\sqrt2} }}} )^2\over r^2}}} = r\sqrt{2-\sqrt{ 2+\sqrt{{ {2+\sqrt2} }}} }. $$ $$\vdots$$
Let $b_n=2^n a_n$. Let $P_n=r\cdot b_n$ be the perimeter of the $2^n$-gon. Let $P$ be the perimeter of the circle. Then $$ \lim_{ n\rightarrow \infty} P_n = P. $$ Note that from the above identity, it follows that the ratio of the perimeter of a circle to its diameter must be a constant, namely $\lim\limits_{n \rightarrow \infty} b_n$. We call this number $\pi$.
Below are some particular calculations when the radius of the circle is $1/2$:
$$\eqalign{ P_2&=2^1\cdot\sqrt 2 \approx 2.82842712\cr P_3&=2^2\cdot\sqrt{2-\sqrt2}\approx 3.06146746\cr P_4&=2^3\cdot\sqrt{2-\sqrt{2+\sqrt2}}\approx3.12144515 \cr P_5&=2^4\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt2}}}\approx 3.13654849\cr P_6&=2^5\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}\approx 3.14033116\cr P_7&=2^6\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}}\approx 3.14127725\cr P_8&=2^7\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}}}\approx 3.1415138 \cr P_9&=2^8\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}}}}\approx 3.14157294 \cr } $$
For completedness:
Remark 1: Here is the proof that the recursion formula (1) holds:
Let $a_n$ be the side length of the $2^n$-gon.
To obtain the $2^{n+1}$-gon: take the "outer end point" of the radii of the circle that bisect the faces of the $2^n$-gon to form the new vertices of the $2^{n+1}$-gon.
We then have, for $a_{n+1}$, the scenario shown in the following diagram (not to scale):
Now $$ b^2=r^2-{a_n^2\over4}; $$ whence $$\eqalign{ a_{n+1}^2={a_n^2\over4} + \Biggl((r-\sqrt{ r^2-{a_n^2\over4}}\ \Biggr)^2 &={a_n^2\over4}+ r^2-2r\sqrt{r^2-{a_n^2\over4}}+r^2 -{a_n^2\over4}\cr &= 2r^2-2r\sqrt{r^2-{a_n^2\over4}}\cr &= 2r^2-r^2\sqrt{4-{a_n^2\over r^2}}\cr &= r^2 \Biggl(2-\sqrt{4-{a_n^2\over r^2}}\ \Biggr).}$$ And, thus $$ a_{n+1}= r \sqrt{2-\sqrt{4-{a_n^2\over r^2}}}. $$
Remark 2: To explain why limit $\lim\limits_{n\rightarrow\infty} P_n=P\ $ holds, I can do no better than refer you to Eric Naslund's comment in his answer.
See also, here.