So you're asking how to prove that the sum of independent chi squares is a chi square with degrees of freedom equal to the sum of the dfs?

There are lots of ways to do that. One way is using the fact that they're independent we know that the MGF (moment generating function) of the sum will be the product of the individual MGFs so then you look at that and recognize it as a the MGF of a chisquare with df equal to the sum of the dfs.