Is the expansion of this expression correct?
Convolution in time domain means product in frequency domain. DTFT of $\delta[n]$ is just $1$, therefore any arbitrary $x[n]$ when convolved with $\delta[n]$ remains same, as in frequency domain you are just multiplying $X(e^{j\omega})$ with $1$.
So, $Y(e^{j\omega}) = 1 \ . \ X(e^{j\omega}) = X(e^{j\omega})$. Hence, $y[n] = x[n]$.
Now, apply same principle for $\delta[n-1]*\delta[n-1]$ to see that the result will be $\delta[n-2]$.