= Solution
Let $Y$ be the number of <isolated vertices> and set $\lambda=e^{-c}$. For fixed $k\geq1$, the <falling factorial> $(Y)_k$ counts ordered $k$-tuples of distinct <isolated vertices>. A specified tuple is isolated exactly when all <edges> incident to its vertices are absent. There are $k(n-k)+\binom k2$ such <edges>, so
$$
\mathbb E(Y)_k=(n)_k(1-p)^{k(n-k)+\binom k2}.
$$
Since $np=\log n+c$ and $np^2=o(1)$, taking <logarithms> gives
$$
\log\mathbb E(Y)_k=k\log n-k(\log n+c)+o(1)=-kc+o(1).
$$
Thus every fixed <factorial moment> tends to $\lambda^k$. We use the standard <factorial-moment criterion for Poisson convergence>: nonnegative integer-valued <random variables> whose <factorial moments> converge, for every fixed order, to $\lambda^k$ converge in distribution to $\operatorname{Poisson}(\lambda)$. Consequently $Y$ has that limiting distribution, and its mass at each fixed integer converges. In particular,
$$
\boxed{\mathbb P(Y\geq1)\longrightarrow1-e^{-e^{-c}}.}
$$
This establishes the limiting <probability> of at least one <isolated vertex>, rather than merely a bound based on its <expectation>.
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