Let $k = \overline{k}$ be a field. We have the inclusion $\iota:k[x,y] \to k[\![x,y]\!]$ and the prime ideal $\mathfrak{p} = (y - \sum_{n\geq 1} x^n/n!)$. This ideal is prime, because it is the kernel of the surjective homomorphism, $\phi : k[\![x,y]\!] \to k[\![x]\!]$ defined by sending $x \mapsto x$ and $y \mapsto \sum_{n\geq 1} x^n/n!$. I want to show that $\iota^{-1}(\mathfrak{p}) = (0)$. For reference, this is II-3 (b) in The Geometry of Schemes by Eisenbud and Harris.
Here is my solution so far. Suppose $f \in k[x,y]$ so that $(\phi\circ\iota)(f) = 0$, i.e. $$0 = f\bigg(x, \sum_{n\geq 1} \frac{x^n}{n!}\bigg).$$ Then, we must show that $f(x,y) = 0$. We can write $f(x,y) = \sum_{i,j} a_{ij}x^iy^j$, where all but finitely many $a_{ij} = 0$, also suppose $deg(f) = d$. Using the fact that the exponential generating function of $(e^x - 1)^j/j! = \sum_{m=0}^\infty S(m,j)x^m/m!$, we get that $$(\phi\circ\iota)(f) = \sum_{m=0}^\infty \sum_{i,j} a_{ij}S(m,j)\frac{j!}{m!}x^{m+i}$$ where $S(m,j)$ are the Stirling numbers of the second kind. Note this power series is zero if and only if the coefficients of $x^n$ are zero for all $n\geq 0$; thus, $$[x^n](\phi\circ\iota)(f) = \sum_{i=0}^n \sum_{j=0}^{n-i} a_{ij} S(n-i,j) \frac{j!}{(n-i)!} = 0.$$ Note that we are using the fact that $S(k,l) = 0$ if $k < l$. Now, for each $n$, we get a linear equation in the $a_{ij}$'s which only involve the coefficients so that $i + j \leq \text{min}(n,d)$. Almost surely, the only solution is the trivial one, because there are infinitely many equations and only $d(d+3)/2$ indeterminates; however, I am having trouble showing this. If you have any ideas on how to finish or maybe try a different approach, it would be very much appreciated!