 Solution

For an explicit Runge-Kutta method, write its strictly lower-triangular stage matrix as $A$, weight vector as $b$, and $1=(1,\dots ,1{)}^T$. Applied to the scalar test equation, its stage vector satisfiesSince $A^s=0$, inversion is a finite Neumann series. Putting $z=\lambda h$ givesThis proves degree at most $s$. The printed statement of degree exactly $s$ needs a nondegeneracy assumption: a two-stage explicit scheme with $b=(1,0)$ is simply Euler's method with an unused second stage, and has polynomial $1+z$ of degree one.