= Solution
The <Poisson distribution> with unit parameter has mass $e^{-1}/j!$. Its <moment-generating function> is
$$
\boxed{\mathbb E[e^{tX}]=e^{-1}\sum_{j=0}^\infty\frac{e^{tj}}{j!}
=\exp(e^t-1)}.
$$
For an integer $k\geq1$, multiply the exponential <probability> bound by $e$ to obtain $\sum_{j=k}^\infty1/j!\leq\exp(e^t-kt)$. The exponent is minimized at $e^t=k$, hence $t=\log k\geq0$, including $t=0$ when $k=1$. Consequently
$$
\boxed{\sum_{j=k}^\infty\frac1{j!}\leq e^{k-k\log k}=\left(\frac ek\right)^k}.
$$
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