Impulse response of forward difference cascaded with one sample delay
For any sequence $x[n]$ you have
$$x[n]\star\delta[n-n_0]=x[n-n_0]\tag{1}$$
because
$$x[n]\star\delta[n-n_0]=\sum_{k=-\infty}^{\infty}x[k]\delta[n-n_0-k]\tag{2}$$
Note that $\delta[n-n_0-k]$ is only non-zero for $k=n-n_0$, hence the result $(1)$.
Now use $x[n]=\delta[n+1]-\delta[n]$ and $n_0=1$.