= Solution
Let $p$ be the <best uniform approximation> to the <exponential function> from the degree-at-most-$n-1$ <polynomial> space. Suppose that $E_{n-1}(e^x)=E_n(e^x)$. Then the same $p$ is best in the larger space. Its error $g(x)=e^x-p(x)$ is nonzero, since the exponential is not a <polynomial>. The <Chebyshev alternation theorem> supplies $n+2$ ordered points where $g$ has alternating signs. By the <intermediate value theorem>, $g$ has at least one zero in each of the $n+1$ intervening intervals, hence at least $n+1$ distinct zeros.
Applying <Rolle's theorem> successively $n$ times forces a zero of $g^{(n)}$. But $\deg p\le n-1$, so
$$
g^{(n)}(x)=e^x>0
$$
everywhere. This is a contradiction. Together with the automatic non-increase of best errors, it proves
$$
\boxed{E_{n-1}(e^x)>E_n(e^x),\qquad n\ge1.}
$$
The argument is an instance of <strict polynomial error decrease from nonvanishing derivatives>.
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