Do the following inductive proof: Let d1, d2, ..., dn, with n at least 2, be positive integers. Use mathematical induction to explain why, if d1 d2 ... dn = 2n-2, then there must be a tree with n vertices whose degrees are exactly d1, d2, ..., dn. (Be careful with reading this statement. It is not the same as saying that any tree with vertex degrees d1, d2, ..., dn must satisfy d1 d2 ... dn = 2n-2, although this is also true. Rather, it says that if you begin with the numbers d1, d2, ..., dn, then you can find such a tree.) Note: For this problem we are concerned with the total degree. Normally in trees we are only concerned with out-degree and usually just say "degree" since the in-degree is always fixed at one in a tree