Let $x, y, z,$ and $w$ be real numbers such that they satisfy the equations:

$$x + y + z + w = 0$$
$$x^7 + y^7 + z^7 + w^7 = 0.$$

We are required to determine the range of the expression $(w + x)(w + y)(w + z)(w)$.

First, note that since $x + y + z + w = 0$, we can express $w$ in terms of $x, y,$ and $z$:

$$w = -(x + y + z).$$

Next, substitute $w$ into the expression $P = (w + x)(w + y)(w + z)(w)$:

$$P = (-(x + y + z) + x)(-(x + y + z) + y)(-(x + y + z) + z)(-(x + y + z)).$$

Simplify each factor:

$$w + x = -(y + z),$$
$$w + y = -(x + z),$$
$$w + z = -(x + y),$$
$$w = -(x + y + z).$$

Then:

$$P = (-(y + z))(-(x + z))(-(x + y))(-(x + y + z)).$$

Since this is a product of four negative terms, $P \geq 0$.

Moreover, from the condition $x^7 + y^7 + z^7 + w^7 = 0$, we know that:

$$x^7 + y^7 + z^7 + (-(x + y + z))^7 = 0.$$

Upon substitution:

$$-(x + y + z)^7 = x^7 + y^7 + z^7.$$

Thus, since all terms are positive by the power and sign symmetry of 7th powers, it implies:

$$x = y = z = w = 0.$$

Substitute back, all are zero, hence:

$$P = 0.$$

Thus, the range of the expression is $\boxed{0}$.