Given that $x$, $y$, and $z$ are positive reals such that $x + y + z = xyz$, we aim to find the minimum value of:
$$x^7(yz-1) + y^7(zx-1) + z^7(xy-1).$$

First, we use the given condition $x + y + z = xyz$. By the AM-GM inequality, we have:
$$xyz = x + y + z \geq 3\sqrt[3]{xyz},$$
which implies:
$$xyz \geq 3\sqrt{3}.$$

Now, consider the given expression:
$$x^7(yz-1) + y^7(zx-1) + z^7(xy-1).$$

Rewriting it, we get:
$$x^7(yz-1) + y^7(zx-1) + z^7(xy-1) = x^6(xyz - x) + y^6(xyz - y) + z^6(xyz - z).$$

This simplifies to:
$$x^6(xyz) + y^6(xyz) + z^6(xyz) - (x^7 + y^7 + z^7).$$

Factoring out $xyz$ from the first part, we get:
$$(x^6 + y^6 + z^6)xyz - (x^7 + y^7 + z^7).$$

Using the given $xyz = x + y + z$, we can rewrite it as:
$$(x^6 + y^6 + z^6)(x + y + z) - (x^7 + y^7 + z^7).$$

Expanding this, we have:
$$(x^7 + y^7 + z^7) + (x^6y + xy^6 + x^6z + xz^6 + y^6z + yz^6) - (x^7 + y^7 + z^7).$$

This simplifies to:
$$x^6y + xy^6 + x^6z + xz^6 + y^6z + yz^6.$$

By the AM-GM inequality, we know:
$$x^6y + xy^6 + x^6z + xz^6 + y^6z + yz^6 \geq 6\sqrt[6]{(xyz)^{14}}.$$

Given $xyz \geq 3\sqrt{3}$, we have:
$$6\sqrt[6]{(3\sqrt{3})^{14}} = 6\sqrt[6]{3^{21}} = 6 \cdot 3^{7/2} = 6 \cdot 27\sqrt{3} = 162\sqrt{3}.$$

Therefore, the minimum value is:
$$\boxed{162\sqrt{3}}.$$

This minimum is achieved when $x = y = z = \sqrt{3}$.