Given is $T(n) = T(n-1) + 2^n$.
After $k$ iterations, the recurrence will look like:
$T(n) = T(n-k) + (2^n + 2^{n-1} + 2^{n-2} + \dots + 2^{n-k+1})$
This recurrence will terminate when $n-k = 1 \implies k = n-1$.
Substituting the same above, we get:
$T(n) = 1 + (2^n + 2^{n-1} + 2^{n-2} + \dots + 2^{2})$ which is a GP with a sum of $\frac{4(2^{n-1} -1)}{2}$ which is $O(2^n)$