Maybe I am misunderstanding your question, you mean you want to count the number of labeled trees on $n$ vertices for which node 1 has degree $k$, right?
Remember that if $f(d_1,d_2,..,d_n)$ is the number of trees in which node $1$ has degree $d_1$ , ..., node $n$ has degree $d_n$ (*) then we have $f(d_1,..,d_n) = \frac{(n-2)!}{(d_1-1)!\cdot .. \cdot (d_n-1)!}$ . This fact can be proved by induction (in the inductive step remember that every tree must have a leaf).
Apply the multinomial theorem to derive a formula for $F(x_1,..,x_n) = \displaystyle\sum_{d_1+d_2+..+d_n = 2\cdot n - 2 ; d_i \geq 1} f(d_1,..,d_n)\cdot x^{d_1}\cdot ..\cdot x^{d_n} $
Here read off the coefficient of $x_1^{k}$ (to force the degree of that vertex) and evaluate it at $x_2=x_3=...=x_n=1$ to get your answer.
(*) We must have $d_i \geq 1$ and $\sum_{i=1}^n d_i = 2\cdot (n - 1)$.