Suppose not. Then $x_{n+T}=x_n$ for all $n$, $T>0$. \[ x_{n+T}=(n+T) \sin x_{n+T-1}+1=(n+T) \sin x_{n-1} +1=n \sin x_{n-1}+1+T\sin x_{n-1}=x_n+T\sin x_{n-1}\]Thus, $T\sin x_{n-1}=0$ for all $n$, so $T=0$ or $\sin x_{n-1}=0$ and $\sin x_{n}=0$, contradiction.