= Solution
For $\operatorname{Im}z>0$, close the $t$-contour in the upper half-plane. <Jordan lemma> removes the large semicircle, and the only enclosed <pole> is the order-$n$ pole at $t=z$. By the <residue at a pole of order n>,
$$
\operatorname{Res}_{t=z}\frac{e^{it}}{(t-z)^n}
=\frac1{(n-1)!}
\left.\frac{d^{\,n-1}}{dt^{\,n-1}}e^{it}\right|_{t=z}
=\frac{i^{n-1}e^{iz}}{(n-1)!}.
$$
The <residue theorem> therefore gives
$$
F(z)=\frac{2\pi i^ne^{iz}}{(n-1)!}.
$$
To continue the integral itself, move the $t$-contour downward locally as $z$ approaches and crosses the real axis, always keeping the pole above the contour. This <analytic continuation by contour deformation> defines
$$
\widetilde F(z)
=\int_{\mathcal C_z}\frac{e^{it}}{(t-z)^n}\,dt,
$$
where $\mathcal C_z$ passes below $z$. Closing $\mathcal C_z$ upward continues to enclose the pole, even when $\operatorname{Im}z\leq0$, so
$$
\boxed{
\widetilde F(z)=\frac{2\pi i^ne^{iz}}{(n-1)!}
}
$$
there as well. The right-hand side is an <entire function>, so it is the unique analytic continuation of $F$ to the whole <complex plane>.
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