Asymptotic behavior of the function $e^{- \lambda t^2}$ when $\lambda$ is small

Question

I wish to prove that when $\lambda$ is taken to be very small $$ \left| e^{- \lambda t^2} - \sum_{n=0}^N \frac{(- \lambda t^2)^n}{n!} \right| = O(e^{-\frac{a}{\lambda}})$$ for some constant $a \in \mathbb{R} : a > 0$. Earlier I proved that $$ \left| e^{- \lambda t^2} - \sum_{n=0}^N \frac{(- \lambda t^2)^n}{n!} \right| \leq \frac{(\lambda t^2)^{N+1}}{(N+1)!}$$ I was given the hint that we may "choose" $N$ such that $N = O(1/\lambda)$, but I am not exactly sure how to use this hint. I am not even sure what $\lambda$ is small means mathematically, nor do I see how I am going to be able to get an exponential decay term with $\lambda$ in it. My first thought was to try and somehow force the factorial of $1/\lambda$ and use the Sterling approximation for $(1/\lambda)!$ but I am unable to evaluate the above inequalities to make use of it.