[proofplan] We verify the Cayley transform by direct algebra. First we check that all displayed rational formulas have non-vanishing denominators on their stated domains, so the maps are holomorphic. Then we compute the defining function $\operatorname{Im}w-|z'|^2$ after applying $C$ and show that it is exactly a positive multiple of $1-|\zeta|^2$. We next prove directly that the proposed inverse formula sends $\mathcal U^n$ into $B^{n+1}$, and only after this domain check do we compose the two maps in both orders. [/proofplan]

[step:Check that the Cayley transform is holomorphic on the unit ball] Let \begin{align*} C:B^{n+1}\to\mathbb C^n\times\mathbb C,\qquad C(\zeta',\zeta_{n+1})=\left(\frac{\zeta'}{1+\zeta_{n+1}},i\frac{1-\zeta_{n+1}}{1+\zeta_{n+1}}\right) \end{align*} be the displayed rational map. If $\zeta=(\zeta',\zeta_{n+1})\in B^{n+1}$, then $|\zeta_{n+1}|\le |\zeta|<1$. Therefore $\zeta_{n+1}\ne -1$, so $1+\zeta_{n+1}\ne0$. Each component of $C$ is a quotient of holomorphic functions with non-vanishing denominator on $B^{n+1}$, hence $C$ is holomorphic on $B^{n+1}$. [/step]

[step:Compute the Siegel defining function after applying $C$]Fix $\zeta=(\zeta',\zeta_{n+1})\in B^{n+1}$, and define $(z',w):=C(\zeta',\zeta_{n+1})$. Put $a:=\zeta_{n+1}\in\mathbb C$. Then \begin{align*} z'=\frac{\zeta'}{1+a},\qquad w=i\frac{1-a}{1+a}. \end{align*} Since $\operatorname{Im}(iq)=\operatorname{Re}(q)$ for every $q\in\mathbb C$, we have \begin{align*} \operatorname{Im}w=\operatorname{Re}\left(\frac{1-a}{1+a}\right). \end{align*} Multiplying numerator and denominator by $1+\overline a$ gives \begin{align*} \frac{1-a}{1+a}=\frac{(1-a)(1+\overline a)}{|1+a|^2}. \end{align*} The real part of the numerator is $1-|a|^2$, because $a-\overline a$ is purely imaginary. Hence \begin{align*} \operatorname{Im}w=\frac{1-|a|^2}{|1+a|^2}. \end{align*} Also, \begin{align*} |z'|^2=\frac{|\zeta'|^2}{|1+a|^2}. \end{align*} Subtracting these two identities gives \begin{align*} \operatorname{Im}w-|z'|^2=\frac{1-|a|^2-|\zeta'|^2}{|1+a|^2}=\frac{1-|\zeta|^2}{|1+\zeta_{n+1}|^2}. \end{align*} Since $\zeta\in B^{n+1}$, the numerator $1-|\zeta|^2$ is positive and the denominator is positive. Thus $\operatorname{Im}w>|z'|^2$, so $C(\zeta)\in\mathcal U^n$.[/step]

[guided]The target domain is defined by one real inequality, so the useful quantity to compute is not $w$ alone but the defining expression $\operatorname{Im}w-|z'|^2$. Fix $\zeta=(\zeta',\zeta_{n+1})\in B^{n+1}$ and define $(z',w):=C(\zeta',\zeta_{n+1})$. To simplify notation, set $a:=\zeta_{n+1}$. Then the formula for $C$ says \begin{align*} z'=\frac{\zeta'}{1+a},\qquad w=i\frac{1-a}{1+a}. \end{align*} We first compute $\operatorname{Im}w$. For any complex number $q$, multiplication by $i$ rotates real and imaginary parts, so $\operatorname{Im}(iq)=\operatorname{Re}(q)$. Applying this to $q=(1-a)/(1+a)$ gives \begin{align*} \operatorname{Im}w=\operatorname{Re}\left(\frac{1-a}{1+a}\right). \end{align*} To extract the real part, rationalize the denominator: \begin{align*} \frac{1-a}{1+a}=\frac{(1-a)(1+\overline a)}{|1+a|^2}. \end{align*} Expanding the numerator gives \begin{align*} (1-a)(1+\overline a)=1+\overline a-a-|a|^2. \end{align*} The term $\overline a-a$ is purely imaginary, so its real part is $0$. Therefore \begin{align*} \operatorname{Im}w=\frac{1-|a|^2}{|1+a|^2}. \end{align*} Next, from $z'=\zeta'/(1+a)$, the Euclidean norm on $\mathbb C^n$ gives \begin{align*} |z'|^2=\frac{|\zeta'|^2}{|1+a|^2}. \end{align*} Subtracting the two displayed formulas yields \begin{align*} \operatorname{Im}w-|z'|^2=\frac{1-|a|^2-|\zeta'|^2}{|1+a|^2}. \end{align*} Since $a=\zeta_{n+1}$ and $|\zeta|^2=|\zeta'|^2+|\zeta_{n+1}|^2$, this becomes \begin{align*} \operatorname{Im}w-|z'|^2=\frac{1-|\zeta|^2}{|1+\zeta_{n+1}|^2}. \end{align*} The denominator is positive because $1+\zeta_{n+1}\ne0$, and the numerator is positive because $\zeta\in B^{n+1}$. Hence $\operatorname{Im}w>|z'|^2$, which is exactly the condition that $C(\zeta)\in\mathcal U^n$.[/guided]

[step:Check that the proposed inverse is holomorphic on the Siegel domain] Define \begin{align*} D:\mathcal U^n\to\mathbb C^n\times\mathbb C,\qquad D(z',w)=\left(\frac{2iz'}{w+i},\frac{i-w}{w+i}\right). \end{align*} If $(z',w)\in\mathcal U^n$, then $\operatorname{Im}w>|z'|^2\ge0$. In particular $\operatorname{Im}w>0$, so $w\ne -i$. Hence $w+i\ne0$ on $\mathcal U^n$. Each component of $D$ is therefore a quotient of holomorphic functions with non-vanishing denominator on $\mathcal U^n$, so $D$ is holomorphic on $\mathcal U^n$. [/step]

[step:Show that the proposed inverse maps the Siegel domain into the ball]Fix $(z',w)\in\mathcal U^n$, and define $(\eta',b):=D(z',w)$. Thus \begin{align*} \eta'=\frac{2iz'}{w+i},\qquad b=\frac{i-w}{w+i}. \end{align*} Since $\operatorname{Im}w>0$, we have $w+i\ne0$. We compute \begin{align*} 1-|\eta'|^2-|b|^2=1-\frac{4|z'|^2}{|w+i|^2}-\frac{|i-w|^2}{|w+i|^2}. \end{align*} Using $|w+i|^2-|i-w|^2=4\operatorname{Im}w$, this becomes \begin{align*} 1-|\eta'|^2-|b|^2=\frac{4(\operatorname{Im}w-|z'|^2)}{|w+i|^2}. \end{align*} Because $(z',w)\in\mathcal U^n$, the numerator is positive and the denominator is positive. Hence $|\eta'|^2+|b|^2<1$, so $D(z',w)=(\eta',b)\in B^{n+1}$.[/step]

[guided]Before composing $C$ with $D$, we must prove that the output of $D$ actually lies in the domain of $C$. Fix $(z',w)\in\mathcal U^n$, and define $(\eta',b):=D(z',w)$, so \begin{align*} \eta'=\frac{2iz'}{w+i},\qquad b=\frac{i-w}{w+i}. \end{align*} The condition $(z',w)\in\mathcal U^n$ gives $\operatorname{Im}w>|z'|^2\ge0$, hence $\operatorname{Im}w>0$ and $w+i\ne0$. Therefore the displayed formula is defined. To show $(\eta',b)\in B^{n+1}$, we compute its squared norm. By the Euclidean norm on $\mathbb C^n\times\mathbb C$, \begin{align*} |\eta'|^2+|b|^2=\frac{4|z'|^2}{|w+i|^2}+\frac{|i-w|^2}{|w+i|^2}. \end{align*} Thus \begin{align*} 1-|\eta'|^2-|b|^2=1-\frac{4|z'|^2}{|w+i|^2}-\frac{|i-w|^2}{|w+i|^2}. \end{align*} Putting the terms over the common positive denominator $|w+i|^2$ gives \begin{align*} 1-|\eta'|^2-|b|^2=\frac{|w+i|^2-|i-w|^2-4|z'|^2}{|w+i|^2}. \end{align*} For a complex number $w=u+iv$, with $u=\operatorname{Re}w$ and $v=\operatorname{Im}w$, we have $|w+i|^2=u^2+(v+1)^2$ and $|i-w|^2=u^2+(1-v)^2$. Hence $|w+i|^2-|i-w|^2=4v=4\operatorname{Im}w$. Therefore \begin{align*} 1-|\eta'|^2-|b|^2=\frac{4(\operatorname{Im}w-|z'|^2)}{|w+i|^2}. \end{align*} The numerator is positive because $\operatorname{Im}w>|z'|^2$, and the denominator is positive because $w+i\ne0$. Hence $1-|\eta'|^2-|b|^2>0$, which is exactly $|\eta'|^2+|b|^2<1$. Therefore $D(z',w)\in B^{n+1}$.[/guided]

[step:Verify that the two formulas are mutual inverses] First fix $\zeta=(\zeta',a)\in B^{n+1}$, where $a:=\zeta_{n+1}$. Let $(z',w):=C(\zeta',a)$. Then \begin{align*} w+i=i\frac{1-a}{1+a}+i=\frac{2i}{1+a}. \end{align*} Therefore the first component of $D(C(\zeta))$ is \begin{align*} \frac{2iz'}{w+i}=\frac{2i\zeta'/(1+a)}{2i/(1+a)}=\zeta', \end{align*} and the second component is \begin{align*} \frac{i-w}{w+i}=\frac{i-i(1-a)/(1+a)}{2i/(1+a)}=a. \end{align*} Thus $D\circ C=\operatorname{id}_{B^{n+1}}$. Conversely fix $(z',w)\in\mathcal U^n$ and set \begin{align*} (\eta',b):=D(z',w)=\left(\frac{2iz'}{w+i},\frac{i-w}{w+i}\right). \end{align*} By the previous step, $(\eta',b)\in B^{n+1}$, so $C(D(z',w))$ is a genuine composition of the stated maps. Then \begin{align*} 1+b=1+\frac{i-w}{w+i}=\frac{2i}{w+i}. \end{align*} The first component of $C(D(z',w))$ is \begin{align*} \frac{\eta'}{1+b}=\frac{2iz'/(w+i)}{2i/(w+i)}=z', \end{align*} and the second component is \begin{align*} i\frac{1-b}{1+b}=i\frac{1-(i-w)/(w+i)}{2i/(w+i)}=w. \end{align*} Thus $C\circ D=\operatorname{id}_{\mathcal U^n}$. [/step]

[step:Conclude that the Cayley transform is a biholomorphism onto the Siegel domain] The previous steps show that $C:B^{n+1}\to\mathcal U^n$ and $D:\mathcal U^n\to B^{n+1}$ are holomorphic, and that $D\circ C=\operatorname{id}_{B^{n+1}}$ and $C\circ D=\operatorname{id}_{\mathcal U^n}$. Hence $D:\mathcal U^n\to B^{n+1}$ is the inverse map of $C$. Therefore $C$ is a biholomorphism from $B^{n+1}$ onto $\mathcal U^n$, with inverse \begin{align*} C^{-1}(z',w)=\left(\frac{2iz'}{w+i},\frac{i-w}{w+i}\right). \end{align*} [/step]