Probability a random node is a leaf node How can I show that in the limit as $n$ approaches $\infty$, the probability of a given node of a tree containing $n$ vertices and $n - 1$ edges being a leaf node approaches $1/e$?

I think the problem I'm having with this is trying to get the right approach to take. I've been trying to write down a limit in terms of $n$ and using the fact that $\lim_{n\to \infty} (1 - 1/n)^{n} = e^{-1},$ but I can't quite write down the probability.
I tried relating the probability for a tree on $n$ nodes to a tree on $n - 1$ nodes (recurrence relation), but I wasn't able to do so. I also tried coming up with a relation for the number of vertices with degree $1$ with the Handshaking Lemma, but this took me nowhere.
Any help is appreciated
 A: Looking at the limit value $1/e$, I think that here we are talking about labeled trees (OP needs to clarify this point!).
By generalizing the classic Cayley's formula $n^{n-2}$ which gives the number of trees on $n$ labeled vertices, we obtain the generating function
$$(x_1+x_2+\dots+x_n)^{n-2}=\sum \prod_{k=1}^n x_k^{d_T(v_k)-1}$$
where, on the right side, each monomial represents a tree $T$ with $n$ labeled vertices $v_k$ each with degree $d_T(v_k)$. Now, a given node, say $v_n$, is a leaf if and only if its  degree is $1$, that is $d_T(v_n)-1=0$. Hence the total number of trees with $n$ labeled vertices where $v_n$ is a leaf can be obtained by letting $x_n=0$, and the other $x_k=1$ in the above formula which yields $(n-1)^{n-2}$.
Hence the desired probability for trees with $n$ labeled nodes is
$$p_n=\frac{(n-1)^{n-2}}{n^{n-2}}=\left(1-\frac{1}{n}\right)^{n-2}\to \frac{1}{e}\quad \text{as $n\to\infty$.}$$
Example. For $n=4$ we have $4^2=16$ labeled trees and
$$p_4=\frac{\text{number of trees with a blue leaf}}{16}=\frac{9}{16}.$$

