How to show that this filter is a HP filter
As was pointed out in the comments above, this is a highpass filter. There is no strict definition for what constitutes a lowpass, bandpass, or highpass filter. Instead, they possess certain qualities:


Lowpass filters attenuate high frequencies while passing low frequencies.
Bandpass filters attenuate low and high frequencies, while passing some band in the middle.
Highpass filters attenuate low frequencies while passing high frequencies.


Note the lack of precise terms in the descriptions above. In order to classify a filter into a category like those above, one evaluates the character of its magnitude transfer function. That requires a little manipulation:

$$
H(\omega) = 1 - e^{-j\omega}
$$

$$
\begin{align}
|H(\omega)| &amp;= |1 - e^{-j\omega}| \\
&amp;= \left|e^{-j\frac{\omega}{2}} \left(e^{j\frac{\omega}{2}} - e^{j\frac{\omega}{2}}\right)\right|
\end{align}
$$

Using Euler's identity, we can simplify the expression inside the parentheses:

$$
e^{j\frac{\omega}{2}} - e^{j\frac{\omega}{2}} = j2\sin\left(\frac{\omega}{2}\right)
$$

$$
|H(\omega)| = \left|e^{-j\frac{\omega}{2}} j2\sin\left(\frac{\omega}{2}\right)\right|
$$

We can drop the $j$ and $e^{-j\frac{\omega}{2}}$ terms here, because each of them have unity magnitude and therefore don't affect $|H(\omega)|$. Therefore:

$$
|H(\omega)| = \left|2\sin\left(\frac{\omega}{2}\right)\right|
$$

We can check this result by comparing it with a plot from MATLAB's Filter Visualization Tool (fvtool):



Which looks about right. This is plotted over the region $\omega \in [0, \pi]$, and shows that the filter has zero response at DC (attenuating low frequencies) and a magnitude response of $2$ at $\omega = \pi$ (passing high frequencies). That looks like a highpass filter.