[proofplan] We use only the defining power-series expansion of the [probability generating function](/page/Probability%20Generating%20Function). First we name the probability masses $p_k=\mathbb P(X=k)$ and observe that the coefficient of $z^n$ in the defining series is exactly $p_n$. If $G_X$ is viewed analytically on the open unit disk, the Taylor coefficient formula at $0$ identifies the same coefficient with $G_X^{(n)}(0)/n!$. [/proofplan] [step:Read the probability mass from the defining power series] For each $k\in\{0,1,2,\dots\}$, define \begin{align*} p_k:=\mathbb P(X=k). \end{align*} By the definition of the probability [generating function](/page/Generating%20Function), \begin{align*} G_X(z)=\sum_{k=0}^{\infty}p_k z^k. \end{align*} The notation $[z^n]G_X(z)$ denotes the coefficient of $z^n$ in this [power series](/page/Power%20Series). Therefore the coefficient of $z^n$ is $p_n$, and hence \begin{align*} [z^n]G_X(z)=p_n=\mathbb P(X=n). \end{align*} [guided] For each $k\in\{0,1,2,\dots\}$, define the probability mass \begin{align*} p_k:=\mathbb P(X=k). \end{align*} The hypothesis that $X$ takes values in $\{0,1,2,\dots\}$ is exactly what makes the probability generating function a power series indexed by the nonnegative integers. By definition of the probability generating function, \begin{align*} G_X(z)=\sum_{k=0}^{\infty}p_k z^k. \end{align*} Now fix $n\in\{0,1,2,\dots\}$. The coefficient-extraction notation $[z^n]G_X(z)$ means: take the coefficient multiplying $z^n$ in the displayed power series for $G_X$. In that series, the term with exponent $n$ is precisely $p_n z^n$. Therefore \begin{align*} [z^n]G_X(z)=p_n. \end{align*} Substituting back the definition of $p_n$ gives \begin{align*} [z^n]G_X(z)=\mathbb P(X=n). \end{align*} [/guided] [/step] [step:Apply the Taylor coefficient formula at the origin] Assume now that $G_X$ is regarded as an [analytic function](/page/Analytic%20Function) \begin{align*} G_X:\mathbb D\to\mathbb C. \end{align*} Since $0\in\mathbb D$, the Taylor coefficient formula for analytic functions at $0$ gives, for every $n\in\{0,1,2,\dots\}$, \begin{align*} [z^n]G_X(z)=\frac{G_X^{(n)}(0)}{n!}. \end{align*} Combining this identity with the coefficient recovery already proved yields \begin{align*} \mathbb P(X=n)=\frac{G_X^{(n)}(0)}{n!}. \end{align*} This proves both asserted formulas. [/step]