= Solution
Let $v_0$ be the highest vector of the <Verma module> $M(n)$ and set $v_k=f^kv_0$. The <Poincaré-Birkhoff-Witt theorem> implies that $(v_k)_{k\geq0}$ is a <basis>, so none of these vectors vanishes. The actions are
$$
f v_k=v_{k+1},\qquad h v_k=(n-2k)v_k,\qquad e v_k=k(n-k+1)v_{k-1}.
$$
The raising operator is a <locally nilpotent operator>: on every vector, repeated application of $e$ eventually removes all its finitely many basis terms. Thus its <algebraic exponential of a locally nilpotent operator> is defined, and $\exp(e)v_0=v_0$. But the middle factor in the proposed product would have to send this vector to
$$
\exp(-f)v_0=\sum_{k\geq0}\frac{(-1)^k}{k!}v_k,
$$
which has infinitely many nonzero, linearly independent basis coefficients. An element of the algebraic <Verma module> is a finite linear combination of the $v_k$, so this expression is not in $M(n)$. In particular the lowering operator is not locally nilpotent.
Even for integral $n\geq0$, $f^{n+1}v_0$ is nonzero in the <Verma module>; it becomes zero only in its finite-dimensional irreducible quotient $L(n)$. Hence \b[$s$ is not defined as the displayed composition of endomorphisms of the algebraic Verma module]. Working in a completion would be a different problem and is not part of the module specified here.
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