Quite right, that was a silly mistake on my part. I did say the smallest element was placed first, but wrote "the list is ultimately sorted in decreasing, not increasing order" backwards.
I meant to contradict the top post: "the largest number is at position 1...we can treat n+1 as a new array of length N-n and solve using the base case proof", which would result in decreasing order (largest first).
The largest number is at position 1 after the first pass of the inner j loop. In fact it's at the ith position after each completion of the j loop, and it serves to separate the sorted and unsorted parts of the array.
Beyond that it's just inserting the ith element into the list by finding its place and shifting everything over one position.
I meant to contradict the top post: "the largest number is at position 1...we can treat n+1 as a new array of length N-n and solve using the base case proof", which would result in decreasing order (largest first).