Solution
= Solution
The assertion is clear for $T_1=x$. If $T_n$ has degree $n$ and leading coefficient $2^{n-1}$, the recurrence shows that the leading term of $T_{n+1}$ is
$$
2x\cdot2^{n-1}x^n=2^nx^{n+1};
$$
the lower-degree <polynomial> $T_{n-1}$ cannot cancel it. Hence
$$
\boxed{\operatorname{lc}(T_n)=2^{n-1}\quad(n\geq1)}.
$$