Proving that the number of vertices of odd degree in any graph G is even I'm having a bit of a trouble with the below question

Given $G$ is an undirected graph, the degree of a vertex $v$, denoted by $\mathrm{deg}(v)$, in graph $G$ is the number of neighbors of $v$.
  Prove that the number of vertices of odd degree in any graph $G$ is even.


Simply, sum of even numbers of odd number is an even number (always odd+odd=even and even+odd=odd and even+even=even). As the sum of degree of vertices needs to be even number, number of such vertices must be even. Which @Mike has presented very succinctly.
