I wanna show the binomial distribution is a valid distribution that is:
given $$p_k= {n\choose{k}}(1-p)^{n-k}p^k$$ prove
$$\sum_{k=0}^{n} p_k=1$$
I did it using induction.
First the base case is $n=1$ in which $$\sum_{k=0}^{1} p_k=1$$
Then the inductive step, let $$\sum_{k=0}^{n}{n\choose{k}}(1-p)^{n-k}p^k=1$$ hold for specific $n$ show that $$\sum_{k=0}^{n+1} {n+1\choose{k}}(1-p)^{n+1-k}p^k=1$$
What I did is that I made a change of variables so that let $n+1=m$, therefore $$\sum_{k=0}^{n+1} {n+1\choose{k}}(1-p)^{n+1-k}p^k=\sum_{k=0}^{m} {m\choose{k}}(1-p)^{m-k}p^k=1$$
This proof seems too fishy for me. So can someone tell me if it's valid or not, and if not why?
You've made one of the common mistakes in induction. What often happens is that people start with some messy formula involving $n$ and then say "$n$ is just a variable, so I can substitute $n+1$ for $n$ and then I have some messy formula involving $n+1$ and I'm done!"
The way that I like to avoid this error, is that you make your inductive hypothesis different. You assume that the claim is true when $n=m$ and use that statement to prove the statement of interest in the case when $n=m+1$. In this set-up, you must think of $m$ as fixed. This prevents you from confusing when $n$ is variable and when $n$ is fixed.
In this problem, Pascal's Identity might be helpful (along with separating a sum, factoring, and a change of variables so that both sums have the same limits).