= Solution
For $\operatorname{Re}z>0$, <integration by parts> gives
$$
\Gamma(z+1)=\int_0^\infty t^ze^{-t}dt=\left[-t^ze^{-t}\right]_0^\infty+z\int_0^\infty t^{z-1}e^{-t}dt.
$$
The boundary term vanishes because $t^{\operatorname{Re}z}\to0$ at zero and exponential decay dominates any fixed power at infinity. Thus
$$
\boxed{\Gamma(z+1)=z\Gamma(z),\qquad \Gamma(1)=\int_0^\infty e^{-t}dt=1.}
$$
This also gives $\Gamma(n+1)=n!$ for nonnegative integers, linking the <Gamma function> to the <factorial>.
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