## ABSTRACT

The Stirling numbers can be used to calculate sums of powers of consecutive integer numbers. Let us first recall the well-known formulas from elementary mathematics: 1 + 2 + 3 + ⋯ + n = 1 2 ( n 2 + n ) . https://s3-euw1-ap-pe-df-pch-content-public-u.s3.eu-west-1.amazonaws.com/9781315122656/55880c27-d289-4466-811a-19fe156133f9/content/umath5_1.jpg"/> The sum of the squares of the first n numbers is 1 2 + 2 2 + 3 2 + ⋯ + n 2 = 1 6 ( 2 n 3 + 3 n 2 + n ) . https://s3-euw1-ap-pe-df-pch-content-public-u.s3.eu-west-1.amazonaws.com/9781315122656/55880c27-d289-4466-811a-19fe156133f9/content/umath5_2.jpg"/> For the third powers we have that 1 3 + 2 3 + 3 3 + ⋯ + n 3 = 1 4 ( n 4 + 2 n 3 + n 2 ) . https://s3-euw1-ap-pe-df-pch-content-public-u.s3.eu-west-1.amazonaws.com/9781315122656/55880c27-d289-4466-811a-19fe156133f9/content/umath5_3.jpg"/> We can easily recognize a rule: the sums of the powers of the first n positive integers can be expressed as polynomials of n polynomial . It would be interesting to know what are the coefficients of these polynomials in general. The sums on the left-hand sides will be called power sums power sum .