= Solution
Put $\alpha_j=1-\beta_j$, so the interior <angle> at the $j$th <polygon> vertex is $\pi\alpha_j$. A <conformal map> straightens that corner by a power: locally,
$$
f(z)-f(x_j)\sim A_j(z-x_j)^{\alpha_j},\qquad f'(z)\sim\alpha_jA_j(z-x_j)^{-\beta_j}.
$$
This follows by composing with the power $1/\alpha_j$ that maps the corner to a half-plane and using <Schwarz reflection> along its straight edges. The reflected map has a nonzero first <derivative> in the straightened coordinates.
For an upper-half-plane <branch of the complex logarithm>, define
$$
P(z)=\prod_{j=1}^n\exp\bigl(-\beta_j\operatorname{Log}(z-x_j)\bigr).
$$
Along each open interval of the real axis, $P$ has constant <argument>. Crossing $x_j$ from left to right decreases $\arg(z-x_j)$ from $\pi$ to zero, so the <argument> of $P$ increases by $\beta_j\pi$. This is exactly the exterior turning <angle> of the polygon <boundary>. Thus $f'/P$ has one constant <boundary> <argument> on all the intervals. The corner asymptotics remove its apparent <singularity> at each <prevertex>.
For a bounded polygon, the sum of exterior <angles> gives $\sum_j\beta_j=2$. Since every vertex has a finite <prevertex>, <infinity> maps to an ordinary point of an edge. In the local coordinate $1/z$, <Schwarz reflection> gives $f(z)=f(\infty)+d/z+O(z^{-2})$ with $d\ne0$. Hence $f'/P$ has a finite nonzero limit at <infinity>. After multiplication by one unit <complex number>, its <boundary values> are real. Reflecting across the real axis extends it to an <entire function>, including each removed <prevertex>, and it is bounded at <infinity>. <Liouville's theorem> makes it constant. Therefore the <Schwarz-Christoffel formula> is
$$
\boxed{f'(z)=c\prod_{j=1}^n(z-x_j)^{-\beta_j},\qquad c\ne0.}
$$
Integration introduces one further additive constant.
The <Möbius transformations> preserving the <upper half-plane> have three real parameters:
$$
z\longmapsto\frac{az+b}{cz+d},\qquad a,b,c,d\in\mathbb R,\quad ad-bc>0,
$$
with common nonzero scaling irrelevant. They can send any cyclically ordered triple of <boundary> points to another such triple. Consequently \b[three prevertices may be fixed independently]; the other $n-3$ are constrained by the polygon's shape, through its side-length ratios. The constants of integration and multiplication account for translation, rotation and scale of the image, rather than further <prevertex> freedom.
If the last <prevertex> is <infinity>, use $1/z$ as its local coordinate. Its factor is absorbed into the overall normalization, leaving
$$
\boxed{f'(z)=c\prod_{j=1}^{n-1}(z-x_j)^{-\beta_j}.}
$$
For a finite image vertex at <infinity>'s preimage, this has asymptotic order $z^{\beta_n-2}$, because $f(z)-f(\infty)$ is of order $z^{-(1-\beta_n)}$. Thus the omitted factor is not evaluated by substituting <infinity> into the finite product; its <angle> is encoded by the behavior at <infinity>.
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