= Solution
For every closed piecewise smooth curve $\gamma$ in $U$, the <winding number> of $f\circ\gamma$ about zero is the integer
$$
m_\gamma=\frac1{2\pi i}\int_\gamma\frac{f'(z)}{f(z)}\,dz.
$$
If $f=g_k^k$ for a holomorphic $k$th root, $g_k$ is nowhere zero, and
$$
m_\gamma=k\frac1{2\pi i}\int_\gamma\frac{g_k'}{g_k}\,dz.
$$
Thus $m_\gamma$ is divisible by every $k$ for which a root exists. An infinite subset of the positive integers is unbounded, so a fixed integer divisible by all those $k$ must be zero. Hence \b[all periods of the holomorphic function $f'/f$ vanish].
Fix $z_0\in U$. Since a domain is connected and open, its points can be joined by piecewise smooth paths. Vanishing periods make
$$
H(z)=\int_{z_0}^{z}\frac{f'(w)}{f(w)}\,dw
$$
independent of path. It is a <holomorphic primitive> with $H'=f'/f$. Differentiating gives $(fe^{-H})'=0$, so $fe^{-H}=f(z_0)$. Choose a complex number $c$ with $e^c=f(z_0)$. Then
$$
\boxed{F=H+c\text{ is holomorphic on }U,\qquad f=e^F.}
$$
This constructs a global <holomorphic logarithm> without requiring $U$ to be simply connected.
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