Inductive hypothesis: When n=k, the conclusion holds.

When n=k+ 1

T has k+ 1 vertices and k edges, and T is connected.

If the degree of each vertex is greater than or equal to 2 (that is, the number of edges drawn from the vertex)

Then the total degree is more than or equal to 2k+2, that is, the number of edges >; =k+ 1 (because each edge has two vertices, that is, it is calculated twice) This is impossible.

So there must be a vertex with a degree less than or equal to 1

And t is a connected graph, that is, the degree of each vertex is greater than or equal to 1.

That is, the degree of vertex A is 1, and this side is AB.

Consider the graph T' with vertex A and edge AB removed,

Note that G' is a graph after removing vertex A and all the edges connected with it, then

T' is a connected subgraph of G', and T' has k vertices and k- 1 edges.

According to the inductive hypothesis, T' is the spanning tree of G'

So T is the spanning tree of G, and the conclusion holds.

Therefore, for any positive integer n, the conclusion holds.