One of my favorite proofs that the sums of each row are powers of two comes from the fact that the numbers in row n+1 are the coefficients of the powers of (a+b)ⁿ, so setting a=b=1 you get 2ⁿ (most discrete math students seeking to prove this end up reaching for induction which is a heavier proof than this).
I like the argument that every number in the row below is formed by summing two numbers from above. So each number above appears twice below. Hence the sum doubles.
One of my favorite proofs that the sums of each row are powers of two comes from the fact that the numbers in row n+1 are the coefficients of the powers of (a+b)ⁿ, so setting a=b=1 you get 2ⁿ (most discrete math students seeking to prove this end up reaching for induction which is a heavier proof than this).