= Solution
After integrating $x_1,\ldots,x_{N-1}$, the remaining exponent contains $(x_T-x_0)^2/N$. The square-root factors telescope:
$$
\begin{aligned}
&\left(\frac{m}{2\pi i\hbar\epsilon}\right)^{N/2}
\prod_{k=1}^{N-1}
\left(\frac{2\pi i\hbar\epsilon}{m}\frac{k}{k+1}\right)^{1/2}\\
&\hspace{2em}
=\sqrt{\frac{m}{2\pi i\hbar\epsilon}}\,
\sqrt{\frac1N}
=\sqrt{\frac{m}{2\pi i\hbar T}}.
\end{aligned}
$$
Since $N\epsilon=T$, the <free-particle propagator> is
$$
\boxed{K(x_T,x_0;T)=
\sqrt{\frac{m}{2\pi i\hbar T}}\,
\exp\left[\frac{im(x_T-x_0)^2}{2\hbar T}\right].}
$$
For $T>0$, the square root has phase $e^{-i\pi/4}$. Its short-time limit is the <position> delta function in the distributional sense, consistent with unitary evolution.
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