Let $T$ be a labeled tree on the set of vertices $\{1,...,n\}$, and its sequence of degrees is $d_1,...,d_n$. Prove that for all $1\le i \le n$ the number of appearances in $F(T)$ (Prufer sequence) is $d_i-1$.
My attempt:
Using induction, the base case is $d_i=1$ this is a leaf so it isn't written to the sequence, only its neighbour.
Assume $d_i=n$ has $n-1$ appearances, and prove that $d_i=n+1$ has $n$ appearances.
Since $d_i$ has $n+1$ neighbours we can write it at most $n+1$ times but since it has the biggest degree, we stop writing when there are two vertices left, which means a single neighbour to $d_i$ is left, thus there are $n$ appearances of $d_i$.
I haven't used the induction hypothesis so I'm suspecting I did something wrong, it also looks too short.
You don't need induction; just analyze the algorithm that generates the Prüfer sequence. Whether vertex $i$ is removed during the algorithm or is one of the two vertices left over at the end of the algorithm, in either case $d_i-1$ of its neigbours have been removed at that point, each resulting in an appearance of $i$ in the Prüfer sequence, and no further appearances of $i$ in the sequence will be generated afterwards.