You can do it without using Master's theorem.
Just do repeated substitutions.
After $k$ iterations, your recurrence will be:
$T(n) = 2^k\times T(n-k) + n(1 + 2 + 4 + \dots + 2^k)$.
The base case for this is satisfied when $n - k = 1$ or $k = n -1$.
You can substitute this in the original equation to get the answer in terms of $n$.