Let $f(\xi)=e^{\pi i \lambda^{-1}|\xi|^2}$ and $$ f_N(\xi)=\sum_{j=0}^N\frac{(\pi i\lambda^{-1}|\xi|^2)^j}{j!}. $$ Show that there exists $C>0$ such that for all $\lambda >1$ and $\xi\in\mathbf{R}^d$: $$ |f(\xi)-f_N(\xi)|\leq C\frac{|\xi|^{2(N+1)}}{\lambda^{N+1}}. $$
This question is motivated by an attempt to understand a step in Wolff's notes on the stationary phase method. If one only needs the estimate uniformly for $\xi$ near zero, then I can see that why this is true since the higher order term of $|\xi|^{2m}$ ($m>N$) is dominated by $|\xi|^{2N+2}$. But I don't see how this is true for all $\xi\in\mathbf{R}^d$.
You can show this by using the integral form of the remainder in Taylor's theorem: $$ e^{ix} = \sum_{j=0}^N\frac{(ix)^j}{j!} + \frac{i^{N+1}}{N!}\int_0^x(x-s)^Ne^{is}\,ds. $$ From this it follows that $$ \bigg|e^{ix}-\sum_{j=0}^N\frac{(ix)^j}{j!}\bigg| \le \frac{|x|^{N+1}}{(N+1)!}. $$ Now substituting $x = \pi\lambda^{-1}|\xi|^2$ shows that the desired inequality holds for all $\xi\in\mathbf R^d$ and all $\lambda > 0$.