= Solution
Let $A,b$ be the coefficients of the $\nu$-stage <Runge-Kutta method>. On the <Dahlquist test equation> $y'=\lambda y$, its <stability function> is
$$
R(z)=1+zb^T(I-zA)^{-1}\mathbf1
=\frac{\det(I-zA+z\mathbf1b^T)}{\det(I-zA)},\qquad z=h\lambda.
$$
Both polynomials have degree at most $\nu$, with denominator normalized to one at zero. Order $2\nu$ implies $R(z)-e^z=O(z^{2\nu+1})$. A rational function with these degree bounds and this accuracy is uniquely the diagonal <Padé approximant> $[\nu/\nu]_{e^z}$. To see uniqueness directly, if $P/Q$ and $\widetilde P/\widetilde Q$ have the required accuracy, then $P\widetilde Q-\widetilde P Q$ is a polynomial of degree at most $2\nu$ vanishing to order at least $2\nu+1$ at zero, hence identically zero.
The diagonal exponential <Padé approximant> can be written
$$
R(z)=\frac{P_\nu(z)}{P_\nu(-z)},\qquad
P_\nu(z)=\sum_{k=0}^{\nu}\frac{(2\nu-k)!\,\nu!}{(2\nu)!\,k!\,(\nu-k)!}z^k.
$$
Use the permitted Padé property that the zeros of $P_\nu(-z)$ lie strictly in the right half-plane and numerator and denominator are coprime. Thus $R$ has no poles in the closed left half-plane. Its real coefficients give $P_\nu(-iy)=\overline{P_\nu(iy)}$, so $|R(iy)|=1$. Also $R(z)\to(-1)^\nu$ at infinity. Applying the <maximum modulus principle> on left half-disks and letting their radii increase proves $|R(z)|\le1$ for $\operatorname{Re}z\le0$. There are no hidden stage poles there either: the irreducible Padé denominator has degree $\nu$, exhausting the possible degree of $\det(I-zA)$, so the latter cannot contain an additional canceled factor. \b[Every such maximal-order Runge-Kutta method is A-stable.] This is <maximal-order Runge-Kutta methods are A-stable>; it does not assert <L-stability>, since the stability function does not tend to zero at infinity.
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