(1) First prove that the proposition holds when n= 1. In practice, just put n= 1 in it, just like letting you prove that "when n+ 1, 1+n=2 holds".

(2) Assume that the proposition holds when n=k, and prove that the proposition holds when n=k+ 1.

You can understand it this way: the first part proves that n= 1 In most propositions, it is true that n takes any nonzero natural number. In this case, let's prove the most basic n= 1.

In the second part, since n=k+1holds when n = k holds, it has been proved that n= 1 holds, and when n= 1+ 1 also holds, that is, n=2. N=2 holds, according to the agreement, n=2+ 1, that is, n=3 holds. Traditionally, n=3+ 1, n = 4+ 1 ... will hold, so all natural numbers can hold propositions.

You can take the first part as a solid foundation. Since it is true that n takes any natural number (as most propositions do), natural n= 1 is also true. The second part is the domino process. 1 Prove 2, 2 Prove 3, 3 Prove 4 ... Prove all nonzero natural numbers.