Proving Identity: $(\stackrel{m + n}{l}) = (\stackrel{m}{l})(\stackrel{n}{0})$

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Homework Statement


Let l, m, and n be positive integers with l \leq m and l \leq n. Prove the identity.
(\stackrel{m + n}{l}) = (\stackrel{m}{0})(\stackrel{n}{l}) + (\stackrel{m}{1})(\stackrel{n}{l-1})+...+(\stackrel{m}{l})(\stackrel{n}{0})

2. The attempt at a solution
I have no clue, I see proof and my brain goes dead. I thought I could just start writing the definition of the parts out on both sides and maybe something would make sense but it got messy quick and I didn't see any light at the end of that tunnel.
 
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How about a proof by induction? The integers l, m, and n have to be positive, so we'd start from l = 1:

<br /> (\stackrel{m+n}{1}) = (\stackrel{m}{0})(\stackrel{n}{1}) + (\stackrel{m}{1})(\stackrel{n}{0}) = m + n<br />

which plainly works. (It doesn't work for l = 0, but it doesn't have to, under the specified conditions.)

Now assume the proposition

(\stackrel{m + n}{l}) = (\stackrel{m}{0})(\stackrel{n}{l}) + (\stackrel{m}{1})(\stackrel{n}{l-1})+...+(\stackrel{m}{l})(\stackrel{n}{0}) .

What happens when we advance to the case for (\stackrel{m + n}{l+1}) ?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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