Recall that $\csc\theta=\frac{1}{\sin\theta}$. Since $\sin45^\circ=\frac{\sqrt{2}}{2}$, we have $$\csc45^\circ=\frac{1}{\sin45^\circ}=\frac{1}{\frac{\sqrt{2}}{2}}=\boxed{\sqrt{2}}.$$
We have $\csc(45^\circ) = \frac{1}{\sin(45^\circ)}$. Since $\sin(45^\circ) = \frac{1}{\sqrt{2}}$, we have $\csc(45^\circ) = \boxed{\sqrt{2}}$.
