If $(X,\tau)^n$ is Hausdorff, is $(X,\tau)$ also Hausdorff? 
If $(X,\tau)^n$ is Hausdorff, is $(X,\tau)$ also Hausdorff?

I know that product of Hausdorff space is Hausdorff, but I want to know if this weaker converse of it is true. Thanks.

Note that $(X,\tau)$ is homeomorphic to a subspace of $(X,\tau)^n$ by mapping $x \mapsto (x,x_0,x_0,\dots,x_0)$ for some fixed $x_0 \in X$ (ignoring the case $X = \emptyset$ which is even easier). Now use the fact that subspaces of Hausdorff spaces are Hausdorff spaces.
