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Using the construction $R_n = K[t^\frac1n]$, $L_n = \text{Quot}(R_n)$ and $P = \bigcup_{n\in \mathbb{N}}L_N$ one automatically gets that the Puiseux series are a field. Nevertheless they are also an algebraically closed field if $K$ is an algebraically closed field. Why is that?

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The proof of the Newton-Puiseux theorem is a relatively straightforward application of Hensel's Lemma. Below is an excerpt from Abhyankar: Algebraic Geometry for Scientists and Engineers. One can read the full two-page proof in Google Books. Abhyankar is a master in this area and a gifted teacher, so I highly recommend reading his expositions. In particular, see also his beautiful exposition Historical ramblings in algebraic geometry and related algebra, which won both a Lester R. Ford award and a Chauvenet prize.

Note that Newton's method and related successive approximation schemes are essentially special cases of a generalized form of Hensel's Lemma. See my 1996.10.15 sci.math post for references. One can also generalize some of these ideas to certain functional and differential equations, for example see my $\rm\:TaylorSolve\:$ command in $\rm\:Macsyma\:.$

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The Puiseux field over $K$ is algebraically closed iff $K$ is algebraically closed of characteristic zero.

A proof of this result using the structure theory of extensions of local fields can be found as Theorem 15 in these notes. Briefly, the idea is that we have to consider unamified, tamely ramified and wildly ramified extensions in order to get from a complete valued field to its algebraic closure. Say we start with the Laurent series field $K((t))$. Then the residue field is $K$, which is algebraically closed, so there are no proper unramified extensions. Then we come to totally tamely ramified extensions -- which are obtained by taking $n$th roots of uniformizers -- and the Kummer theory comes out especially nicely here (Theorem 11): it turns out that we get precisely to the Puiseux series field $\bigcup_n K((t^{\frac{1}{n}}))$. Now in general (and in particular, in positive characteristic!) we would still have to contend with the wildly ramified extensions. But here the residue characteristic is zero, so by definition there are no wildly ramified extensions. So we've made it all the way to the algebraic closure.

[Other answers are also possible. Perhaps someone will provide one...]

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I would have posted a less clear version of the same explanation, but let me comment to advertise a bit; this result (with the same argument) is also in Theorem 4.18 of chapter 11 of Note that the argument for the strcture theory of totally tamely ramified extensions is quite simple in view of Hensel's lemma. – Akhil Mathew May 24 '11 at 19:54

As Pete Clark pointed out, you need $K$ to be of characteristic zero for this to be valid. To say that the Puiseux series are an algebraically closed field is to say that if $f(x,y) = \sum_{i=0}^n P_i(x)y^i$ is a polynomial whose coefficients are fractional Puiseux series $P_i(x) = \sum_{j = 0}^{\infty} a_{ij} x^{j \over N}$, with $P_n(x) = 1$, then one has a factorization $$f(x,y) = \prod_{i=1}^n (y - Q_i(x))\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(*)$$ Here the "roots" $Q_i(x)$ are also fractional Puiseux series over $K$, possibly with a larger $N$.

The reason this is true is as follows. Newton's original method for finding Puiseux series for polynomials over ${\mathbb R}$ works over arbitrary $K$ of characteristic zero, in the sense that it produces a fractional Puiseux series $Q_1(x)$ such that $f(x,y)$ factors into $(y - Q_1(x))g(x,y)$ where now $g(x,y)$ is of the form $\sum_{i=0}^{n-1} R_i(x)y^i$ where the $R_i(x)$ are fractional Puiseux series. (If you think about it, there's a natural way to multiply $y - Q_1(x)$ and $g(x,y)$ together so this is well-defined.)

And it doesn't work just for polynomials, it works for all $f(x,y)$ of the above form $\sum_{i=0}^n P_i(x)y^i$ as well; $f(x,y)$ doesn't have to be a polynomial, and the presence of fractional powers don't play any major role since one is effectively just replacing the variable $x$ by the variable $x^{1 \over N}$ for some large $N$.

Once you have $f(x,y) = (y - Q_1(x))g(x,y)$ you can then apply the procedure to $g(x,y)$ repeatedly until you have the factorization $(*)$.

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