One way to see this is as follows.
Pick a basis $\{x_1,\dots,x_n\}$ for $V$. Then the set $B$ of all non-commutative monomials $x_{i_1}x_{i_2}\cdots x_{i_r}$ of all lengths $r\geq0$ (including the zero length) and with $i_1,\dots,i_r\in\{1,\dots,n\}$, is a basis of $TV$. Moreover, this elements are multiplied by juxtaposition.
Now suppose $a$ and $b$ are non-zero elements of $TV$. Then we can write both of them as a linear combination of the elements of $B$. Suppose that $x_{i_1}x_{i_2}\cdots x_{i_r}$ is one of the monomials of $B$ which appears in $a$ with a non-zero coefficient and has maximal length, and suppose that $x_{j_1}x_{j_2}\cdots x_{j_s}$ is one of the monomials of $B$ which appears in $b$ with a non-zero coefficient and has maximal length. Then you can show that the monomial $x_{i_1}x_{i_2}\cdots x_{i_r}x_{j_1}x_{j_2}\cdots x_{j_s}$ is a monomials of $B$ which appears in $ab$ with a non-zero coefficient: it follows then that $ab\neq0$.