TITLE: A graph with $n+1$ vertices and a vertex of every degree $1\dots n$
QUESTION [8 upvotes]: Let $G$ be a graph with $n+1$ vertices. Suppose that for every $i=1,\ldots,n$ there is a vertex in $G$ of degree $i$. What is the degree of the other vertex (that is, what degree is repeated)?
I know that if $n$ is even the degree of the other vertex is $\dfrac{n}{2}$. If $n$ is odd the degree of the other vertex is $\dfrac{n+1}{2}$. I get prove supponsing that the repeated degree wasn't $\dfrac{n}{2}$ or $\dfrac{n+1}{2}$, but it is very large. Can I do it in a more compact way?

REPLY [7 votes]: Let $u,v$ be the vertices of degree $n$ and $1$. Then $u$ is connected to all vertices and in particular to $v$.
Then erasing $u$ and $v$ you get a new graph with $n-1$ vertices, where all the other $n-1$ degrees decreased by exactly $1$. Thus, in the new graph the degrees are $1,2,3,..., n-2$ and $??$...
This suggests induction and this is actually the proof of $P(n) \Rightarrow P(n+2)$.
Now: Check the formula for $n=1$ and then by induction $P(n)\Rightarrow P(n+2)$ you prove your formula for all odd integers.
$n=2$ and induction proves it for even ones. This proof moreover kinda explains why you get different formulas for odd/even... 
P.S. Erasing the largest vertex is actually exactly the algorithm from Hakimi-Havel theorem, if you are familiar with it you can try to re-write the proof using that Theorem, but induction is probably clearer and simpler.