= Solution
Write the <Runge-Kutta method> coefficients as $A\in\mathbb R^{\nu\times\nu}$, $b\in\mathbb R^\nu$, and let $e$ be the all-ones column. On the test equation $y'=\lambda y$, with $z=h\lambda$, the stage <vector> satisfies $Y=ey_n+zAY$ and the next value is $y_{n+1}=y_n+zb^TY$. Hence its <stability function> is
$$
R(z)=1+zb^T(I-zA)^{-1}e
=\frac{\det(I-zA+zeb^T)}{\det(I-zA)}.
$$
Both <polynomials> have degree at most $\nu$, and the denominator has value one at zero. <Order of a Runge-Kutta method> $2\nu$ implies that this test-equation approximation agrees with $e^z$ through degree $2\nu$:
$$
R(z)-e^z=O(z^{2\nu+1}).
$$
By uniqueness of the diagonal <Padé approximant>, $R=[\nu/\nu]_{e^z}$. Explicitly,
$$
R(z)=\frac{P_\nu(z)}{P_\nu(-z)},\qquad
P_\nu(z)=\sum_{j=0}^\nu\frac{(2\nu-j)!\,\nu!}{(2\nu)!\,j!\,(\nu-j)!}\,z^j.
$$
The permitted diagonal <Padé approximant> property says this <rational function> has no poles in the closed left half-plane and satisfies $|R(z)|\le1$ there. Therefore
$$
\boxed{\text{Every }\nu\text{-stage Runge-Kutta method of order }2\nu\text{ is A-stable}.}
$$
There is also no hidden stage-solve singularity from a canceled denominator: the <Padé approximant> numerator and denominator are coprime and each has degree $\nu$. The original <determinant> denominator already has degree at most $\nu$, so, with its normalization at zero, it equals $P_\nu(-z)$ and cannot contain an additional canceled factor. This proves <maximal-order Runge-Kutta methods are A-stable>.
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