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.

I think this is true. For any two points $x\neq y$ of $(X,\tau)$, consider the two points
$$
(x,x_2,\dots,x_n)\neq(y,x_2,\dots,x_n)
$$
which have disjoint neighborhoods, say $U,V$. Judging from your notation, the product topology on $(X,\tau)^n$ is implied. So it is generated by basis of the form $W_1\times\cdots W_n$ where $W_i$ are open sets of $X$. By the definition of basis, exists two basis element $U_1\times\cdots\times U_n$ and $V_1\times\cdots\times V_n$ s.t.
$$
\begin{split}
(x,x_2,\dots,x_n)\in U_1\times\cdots\times U_n\subset U \\
(y,x_2,\dots,x_n)\in V_1\times\cdots\times V_n\subset U \\
\end{split}
$$
Since $U,V$ are disjoint:
$$
(U_1\times\cdots\times U_n)\cap(V_1\times\cdots\times V_n)=(U_1\cap V_1)\times\cdots\times (U_n\cap V_n)=\varnothing
$$
It is then obvious that $U_i\cap V_i\neq\varnothing$ for $i\geq 2$, and $U_1\cap V_1$ has to be empty. This is the required disjoint neighborhoods for $x,y$ in $(X,\tau)$.
