 Solution

Insert a smooth exact function into the multistep method and expand its residual about $t_n$. DefineThe residual is ${\sum }_{k\ge 0}{h}^k{c}_k{y}^{(k)}(t_n)/k!$. Therefore order at least $p$ is equivalent to $c_0=\cdots =c_p=0$. Applying the residual to the monomials $(t-t_n{)}^k$, $k\le p$, gives exactly those conditions. Linearity then makes them equivalent to exactness on every polynomial of degree at most $p$, as asserted. For exact rather than at-least order $p$, the next coefficient must not vanish.

If an $s$-step formula were exact for degree $2s+1$, use Hermite interpolation to choose $Q$ with $Q(t_{n+i})=0$ for $i<s$, $Q(t_{n+s})=1$, and $Q^{\prime }(t_{n+i})=0$ for every $i$. Its degree is at most $2s+1$. The purported identity would then read ${\rho }_s=0$, contradicting ${\rho }_s=1$. No $s$-step method has order $2s+1$ or higher.