1. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{n}=2^{n} a_{n-1}(n>1)$, and $a_{1}=1$, the general term formula of the sequence $a_{n}=$ ( ).
A. $2^{\frac{n(n-1)}{2}}$
B. $2^{\frac{n(n+1)}{2}}$
C. $2^{\frac{n^{2}+n-2}{2}}$
D. $2^{\frac{n^{2}-n+2}{2}}$
1. C.

From the conditions, we have $\frac{a_{n}}{a_{n}-1}=2^{n}(n>1), \frac{a_{n-1}}{a_{n-2}}=2^{n-1}, \cdots, \frac{a_{2}}{a_{1}}=2^{2}$. Multiplying them together, we get $\frac{a_{n}}{a_{1}}=2^{2+3+\cdots+n}$, and since $a_{1}=1$, we have $a_{n}=2^{\frac{(n+2)(n-1)}{2}}=2^{\frac{n}{2}^{2}+n-2}$.