Elementary Foundations: An introduction to topics in discrete mathematics

This is just the ordinary Law of Syllogism.

Let $k\ge 3\text{.}$ Consider the $n=k$ version (below left) and the $n=k+1$ version (below right) of the Extended Law of Syllogism.

Assume the $n=k$ version of the argument is valid. We want to show that the $n=k+1$ version is also valid. So suppose that premises of that latter version are all true. We need to show that the conclusion $p_1\to p_{k+1}$ must then also be true.

But each premise of the $n=k$ version is also a premise of the $n=k+1$ version, so we can say that we have assumed that every premise of the $n=k$ version is true. But we have also assumed that version to be valid, so we may take its conclusion $p_1\to p_k$ to be true.

Consider the following syllogism.

Since this is valid (base case $n=2$) and its premises are all true, the conclusion is true.