The argument, first given by Cauchy, hinges on Cauchy's integral formula and the power series expansion of the expression

${\displaystyle {\frac {1}{w-z}}.}$ 

Let ${\displaystyle D}$  be an open disk centered at ${\displaystyle a}$  and suppose ${\displaystyle f}$  is differentiable everywhere within an open neighborhood containing the closure of ${\displaystyle D}$ . Let ${\displaystyle C}$  be the positively oriented (i.e., counterclockwise) circle which is the boundary of ${\displaystyle D}$  and let ${\displaystyle z}$  be a point in ${\displaystyle D}$ . Starting with Cauchy's integral formula, we have

${\displaystyle {\begin{aligned}f(z)&{}={1 \over 2\pi i}\int _{C}{f(w) \over w-z}\,\mathrm {d} w\\[10pt]&{}={1 \over 2\pi i}\int _{C}{f(w) \over (w-a)-(z-a)}\,\mathrm {d} w\\[10pt]&{}={1 \over 2\pi i}\int _{C}{1 \over w-a}\cdot {1 \over 1-{z-a \over w-a}}f(w)\,\mathrm {d} w\\[10pt]&{}={1 \over 2\pi i}\int _{C}{1 \over w-a}\cdot {\sum _{n=0}^{\infty }\left({z-a \over w-a}\right)^{n}}f(w)\,\mathrm {d} w\\[10pt]&{}=\sum _{n=0}^{\infty }{1 \over 2\pi i}\int _{C}{(z-a)^{n} \over (w-a)^{n+1}}f(w)\,\mathrm {d} w.\end{aligned}}}$ 

Interchange of the integral and infinite sum is justified by observing that ${\displaystyle f(w)/(w-a)}$  is bounded on ${\displaystyle C}$  by some positive number ${\displaystyle M}$ , while for all ${\displaystyle w}$  in ${\displaystyle C}$ 

${\displaystyle \left|{\frac {z-a}{w-a}}\right|\leq r<1}$ 

for some positive ${\displaystyle r}$  as well. We therefore have

${\displaystyle \left|{(z-a)^{n} \over (w-a)^{n+1}}f(w)\right|\leq Mr^{n},}$ 

on ${\displaystyle C}$ , and as the Weierstrass M-test shows the series converges uniformly over ${\displaystyle C}$ , the sum and the integral may be interchanged.

As the factor ${\displaystyle (z-a)^{n}}$  does not depend on the variable of integration ${\displaystyle w}$ , it may be factored out to yield

${\displaystyle f(z)=\sum _{n=0}^{\infty }(z-a)^{n}{1 \over 2\pi i}\int _{C}{f(w) \over (w-a)^{n+1}}\,\mathrm {d} w,}$ 

which has the desired form of a power series in ${\displaystyle z}$ :

${\displaystyle f(z)=\sum _{n=0}^{\infty }c_{n}(z-a)^{n}}$ 

with coefficients

${\displaystyle c_{n}={1 \over 2\pi i}\int _{C}{f(w) \over (w-a)^{n+1}}\,\mathrm {d} w.}$ 

• Since power series can be differentiated term-wise, applying the above argument in the reverse direction and the power series expression for $${\displaystyle {\frac {1}{(w-z)^{n+1}}}}$$  gives $${\displaystyle f^{(n)}(a)={n! \over 2\pi i}\int _{C}{f(w) \over (w-a)^{n+1}}\,dw.}$$  This is a Cauchy integral formula for derivatives. Therefore the power series obtained above is the Taylor series of ${\displaystyle f}$ .
• The argument works if ${\displaystyle z}$  is any point that is closer to the center ${\displaystyle a}$  than is any singularity of ${\displaystyle f}$ . Therefore, the radius of convergence of the Taylor series cannot be smaller than the distance from ${\displaystyle a}$  to the nearest singularity (nor can it be larger, since power series have no singularities in the interiors of their circles of convergence).
• A special case of the identity theorem follows from the preceding remark. If two holomorphic functions agree on a (possibly quite small) open neighborhood ${\displaystyle U}$  of ${\displaystyle a}$ , then they coincide on the open disk ${\displaystyle B_{d}(a)}$ , where ${\displaystyle d}$  is the distance from ${\displaystyle a}$  to the nearest singularity.

• "Existence of power series". PlanetMath.