The problem asks us to evaluate the infinite sum $\sum_{n=1}^{\infty} \frac{x^n}{(n-1)!}$.

1. Problem Description

The problem asks us to evaluate the infinite sum $\sum_{n=1}^{\infty} \frac{x^n}{(n-1)!}$.

2. Solution Steps

We can rewrite the sum by changing the index. Let $k = n-1$. Then $n = k+1$. When $n=1$, $k=0$. So we have

$$\sum_{n=1}^{\infty} \frac{x^n}{(n-1)!} = \sum_{k=0}^{\infty} \frac{x^{k+1}}{k!} = x \sum_{k=0}^{\infty} \frac{x^k}{k!}$$

We know that the Taylor series expansion for $e^x$ is

$$e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$

Thus,

$$x \sum_{k=0}^{\infty} \frac{x^k}{k!} = x e^x$$

3. Final Answer

$xe^x$