Let $\mathcal{X}$ be a smooth geometrically irreducible projective curve over $\mathbb{F}_q$. Fix a closed point $\infty\in \mathcal{X}(\bar{\mathbb{F}_q})$. Let $K$ be the function field of $\mathcal{X}$ and $\mathcal{A}$ the ring of functions on $\mathcal{X}$ regular away from $\infty$. Let $\phi: \mathcal{A}\rightarrow \mathcal{F}\{\tau\}$ be a Drinfeld module where $\mathcal{F} $ has generic characteristic. Then $End_{\mathcal{F}}(\phi)$ is a commutative $\mathcal{A}$-module. My question is on the proof of this fact in Goss's book Basic Structure of Function Fields Arithmetic (Proposition 4.7.6):
First of all $\phi$ can be defined over a finitely generated subfield of $\mathcal{F}$ over $K$, say $K'$, this can be done by adjoining the coefficients of finitely many elements of the image of $\phi$. The next claim is that $K'$ can be embedded inside $\mathbb{C}_{\infty}$ where $\mathbb{C}_{\infty}$ is obtained by completing $K$ w.r.t $\infty$ closing and completing again.
So my question is why (how) can we embed $K'$ in $\mathbb{C}_{\infty}$?
Thanks
If we had $k = \mathbb{Q}$ and $K' = \mathbb{Q}(x)$, then we would embed $K'$ into $\mathbb{C}$ by $x \mapsto \pi$ if $x$ was transcendental over $\mathbb{Q}$ and by $x \mapsto \alpha$ if $x$ was a root of the irreducible $f$ and $\alpha \in \mathbb{C}$ is one of the roots of $f$ in $\mathbb{C}$.
So suppose that $K' = k(x_1, x_2, \ldots, x_n)$ and put $K_1 = k(x_1, x_2, \ldots, x_{n - 1})$ and that we have an embedding$$k(x_1, \ldots, x_{n - 1}) \hookrightarrow \mathbb{C}_\infty.$$Suppose $x_n$ is algebraic over $K_1$ and that $f$ is the irreducible polynomial with coefficients in $K_1$ and $x_n$ as a root. Then using the embedding $K_1 \hookrightarrow \mathbb{C}_\infty$, $f$ has coefficients in $\mathbb{C}_\infty$ which is algebraically closed, so choose a root $\alpha$ and then we have an embedding $K_1(x) \hookrightarrow \mathbb{C}_\infty$ by sending $x \mapsto \alpha$.
If $x_n$ is not algebraic over $K_1$, then choose an element $\alpha \in \mathbb{C}_\infty$ which is transcendental over $K_1$. If we can not do this, it would mean that $\mathbb{C}_\infty$ would be an algebraic extension of a finitely generated extension of $k$. So in this case, we get an embedding too.