Inverse Discrete Time Fourier Transform of $1$
The IDTFT of $X(e^{j\omega})=1$ is indeed

$$x[n]=\frac{\sin(n\pi)}{n\pi}\tag{1}$$

Now, what happens for indices $n\neq 0$? As it turns out, you can safely rewrite $(1)$ as

$$x[n]=\delta[n]\tag{2}$$

where $\delta[n]$ is the discrete-time unit impulse. (HINT: think about where the zeros of $\sin(x)$ are).