As has been made clear in the comments, the result is false as stated. However, it is approximately true in the limit as $it/N\to0$ (and $j$ is not too large).
Here is one way to get a rigorous estimate of this nature. First, note that $\ln(1-x)=-x+O(x^2)$ as $x\to0$ (you can easily give rigorous upper bounds on the $O(x^2)$ term if needed). With $x=it/N$, we take exponentials, then raise both sides to the $j$th power to get $\Bigl(1-\frac{it}{N}\Bigr)^j=\exp\biggl(-\frac{ijt}{N}+jO\Bigl(\Bigl(\frac{it}{N}\Bigr)^2\Bigr)\biggr)$ where the big O refers to the limit $it/N\to0$. The missing factor $1-it/N$ on the left hand side (because of $j$ replacing $j+1$) is clearly close to $1$ in this case.