Observe that $F_{X/S}$ is a finite morphism by Varieties, Lemma \ref{varieties-lemma-relative-frobenius-finite}. To prove that $F_{X/S}$ is flat, it suffices to show that the morphism $F_{X/S, s} : X_s \to X^{(p)}_s$ between fibres is flat for all $s \in S$, see More on Morphisms, Theorem \ref{more-morphisms-theorem-criterion-flatness-fibre}. Flatness of $X_s \to X^{(p)}_s$ follows from Algebra, Lemma \ref{algebra-lemma-CM-over-regular-flat} (and the finiteness already shown). By More on Morphisms, Lemma \ref{more-morphisms-lemma-lci-permanence} the morphism $F_{X/S}$ is a local complete intersection morphism. Hence $F_{X/S}$ is finite syntomic (see More on Morphisms, Lemma \ref{more-morphisms-lemma-flat-lci}). \medskip\noindent For every point $x \in X$ we may choose a commutative diagram $$ \xymatrix{ X \ar[d] & U \ar[l] \ar[d]_\pi \\ S & \mathbf{A}^d_S \ar[l] } $$ where $\pi$ is \'etale and $x \in U$ is open in $X$, see Morphisms, Lemma \ref{morphisms-lemma-smooth-etale-over-affine-space}. Observe that $\mathbf{A}^d_S \to \mathbf{A}^d_S$, $(x_1, \ldots, x_d) \mapsto (x_1^p, \ldots, x_d^p)$ is the relative Frobenius for $\mathcal{A}^d_S$ over $S$. The commutative diagram $$ \xymatrix{ U \ar[d]_\pi \ar[r]_{F_{X/S}} & U^{(p)} \ar[d]^{\pi^{(p)}} \\ \mathbf{A}^d_S \ar[r]^{x_i \mapsto x_i^p} & \mathbf{A}^d_S } $$ of Varieties, Lemma \ref{varieties-lemma-relative-frobenius-endomorphism-identity} for $\pi : U \to \mathbf{A}^d_S$ is cartesian by \'Etale Morphisms, Lemma \ref{etale-lemma-relative-frobenius-etale}. Since the construction of $\Theta$ is compatible with base change and since $\Omega_{U/S} = \pi^*\Omega_{\mathbf{A}^d_S/S}$ (Lemma \ref{lemma-etale}) we conclude that it suffices to show the lemma for $\mathbf{A}^d_S$. \medskip\noindent Let $A$ be a ring of characteristic $p$. Consider the unique $A$-algebra homomorphism $A[y_1, \ldots, y_d] \to A[x_1, \ldots, x_d]$ sending $y_i$ to $x_i^p$. The arguments above reduce us to computing the map $$ \Theta^i : \Omega^i_{A[x_1, \ldots, x_d]/A} \to \Omega^i_{A[y_1, \ldots, y_d]/A} $$ We urge the reader to do the computation in this case for themselves. As in Example \ref{example-Garel} we may reduce this to computing a formula for $\Theta^i$ in the universal case $$ \mathbf{Z}[y_1, \ldots, y_d] \to \mathbf{Z}[x_1, \ldots, x_d],\quad y_i \mapsto x_i^p $$ In turn, we can find the formula for $\Theta^i$ by computing in the complex case, i.e., for the $\mathbf{C}$-algebra map $$ \mathbf{C}[y_1, \ldots, y_d] \to \mathbf{C}[x_1, \ldots, x_d],\quad y_i \mapsto x_i^p $$ We may even invert $x_1, \ldots, x_d$ and $y_1, \ldots, y_d$. In this case, we have $\text{d}x_i = p^{-1} x_i^{- p + 1}\text{d}y_i$. Hence we see that \begin{align*} \Theta^i( x_1^{e_1} \ldots x_d^{e_d} \text{d}x_1 \wedge \ldots \wedge \text{d}x_i) & = p^{-i} \Theta^i( x_1^{e_1 - p + 1} \ldots x_i^{e_i - p + 1} x_{i + 1}^{e_{i + 1}} \ldots x_d^{e_d} \text{d}y_1 \wedge \ldots \wedge \text{d}y_i ) \\ & = p^{-i} \text{Trace}(x_1^{e_1 - p + 1} \ldots x_i^{e_i - p + 1} x_{i + 1}^{e_{i + 1}} \ldots x_d^{e_d}) \text{d}y_1 \wedge \ldots \wedge \text{d}y_i \end{align*} by the properties of $\Theta^i$. An elementary computation shows that the trace in the expression above is zero unless $e_1, \ldots, e_i$ are congruent to $-1$ modulo $p$ and $e_{i + 1}, \ldots, e_d$ are divisible by $p$. Moreover, in this case we obtain $$ p^{d - i} y_1^{(e_1 - p + 1)/p} \ldots y_i^{(e_i - p + 1)/p} y_{i + 1}^{e_{i + 1}/p} \ldots y_d^{e_d/p} \text{d}y_1 \wedge \ldots \wedge \text{d}y_i $$ We conclude that we get zero in characteristic $p$ unless $d = i$ and in this case we get every possible $d$-form.