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Steinberg's "Lecture Notes on Chevalley Groups" the Corollary of Lemma 54 on page 132. There is possibly a more modern reference.

EDIT: Sorry the reference to Steinberg is not sufficient as he does not treat arbitrary finite reductive groups. This follows from a theorem of Rosenlicht about the fixed points of a unipotent group under a Frobenius endomorphism. If $\mathbf{G}$ is a connected reductive algebraic group with Frobenius endomorphism $F : \mathbf{G} \to \mathbf{G}$ and $\mathbf{U} \leqslant \mathbf{G}$ is an $F$-stable unipotent subgroup then the finite group $\mathbf{U}^F$ has order $q^{\dim\mathbf{U}}$, (see Geck's book "An Introduction To Algebraic Geometry and Algebraic Groups" - Theorem 4.2.4). Now take $\mathbf{G} = \mathrm{GL}_n(\overline{\mathbb{F}}_p)$ and $F$ to be the map given by $F(x_{ij}) = (x_{ij}^q)$ where $q = p^a$ for some integer $a$ then $\mathbf{G}^F = \mathrm{GL}_n(q)$. The subgroup $\mathbf{B} \leqslant \mathbf{G}$ of upper triangular matrices is an $F$-stable Borel subgroup of $\mathbf{G}$ and its unipotent radical is the subgroup of all upper uni-triangular matrices. This has maximal dimension amongst all unipotent subgroups of $\mathbf{G}$, which can be seen in the following way. Assume $\mathbf{V} \leqslant \mathbf{G}$ is a unipotent subgroup of $\mathbf{G}$ then as $\mathbf{V}$ is solvable it is contained in a Borel subgroup $\mathbf{B}'$ of $\mathbf{G}$. As $\mathbf{V}$ is a unipotent subgroup of $\mathbf{B}'$ it is contained in the unipotent radical of $\mathbf{B}'$. As all Borel subgroups of $\mathbf{G}$ are conjugate we have the unipotent radicals of $\mathbf{B}$ and $\mathbf{B}'$ have the same dimension, hence we have $\dim\mathbf{V} \leqslant \dim \mathbf{U}$. Now applying Rosenlicht's theorem we see that the order of $\mathbf{U}^F$ is a power of $p$. We can now apply the order formula for finite reductive groups, (see Geck - Corollary 4.2.5), and easily deduce that this is the maximal power of $p$ dividing the order of $\mathbf{G}^F$. This approach has the advantage that it works for all finite reductive groups.

Steinberg's "Lecture Notes on Chevalley Groups" the Corollary of Lemma 54 on page 132. There is possibly a more modern reference.

EDIT: Sorry the reference to Steinberg is not sufficient as he does not treat arbitrary finite reductive groups. However all that is needed is to obtain a slightly more general formula for the order of an arbitrary finite reductive group. This is obtained, for instance, in Corollary 4.2.5 of Geck's book "An Introduction to Algebraic Geometry and Algebraic Groups". the important point is a theorem of Rosenlicht about the fixed points of a connected unipotent group under a Frobenius endomorphism. If $\mathbf{U}$ is a connected unipotent affine algebraic group with Frobenius endomorphism $F' : \mathbf{U} \to \mathbf{U}$ then the finite group $\mathbf{U}^{F'}$ has order $q^{\dim\mathbf{U}}$, (see Geck - Theorem 4.2.4). 

Now take $\mathbf{G} = \mathrm{GL}_n(\overline{\mathbb{F}}_p)$ and $F$ to be the map given by $F(x_{ij}) = (x_{ij}^q)$ where $q = p^a$ for some integer $a$ then $\mathbf{G}^F = \mathrm{GL}_n(q)$. The subgroup $\mathbf{B} \leqslant \mathbf{G}$ of upper triangular matrices is an $F$-stable Borel subgroup of $\mathbf{G}$ and its unipotent radical $\mathbf{U} \leqslant \mathbf{B}$ is the subgroup of all upper uni-triangular matrices. This has maximal dimension amongst all connected unipotent subgroups of $\mathbf{G}$, which can be seen in the following way. Assume $\mathbf{V} \leqslant \mathbf{G}$ is a connected unipotent subgroup of $\mathbf{G}$ then as $\mathbf{V}$ is solvable it is contained in a Borel subgroup $\mathbf{B}'$ of $\mathbf{G}$. As $\mathbf{V}$ is a unipotent subgroup of $\mathbf{B}'$ it is contained in the unipotent radical of $\mathbf{B}'$ and as all Borel subgroups of $\mathbf{G}$ are conjugate we have the unipotent radicals of $\mathbf{B}$ and $\mathbf{B}'$ have the same dimension. In particular, we have $\dim\mathbf{V} \leqslant \dim \mathbf{U}$. Now applying Rosenlicht's theorem we see that the order of $\mathbf{U}^F$ is a power of $p$. We can now apply Steinberg's argument to the order formula for finite reductive groups to deduce that the index of $\mathbf{U}^F$ in $\mathbf{G}^F$ is coprime to $p$. This approach has the advantage that it shows that the fixed points of the unipotent radical of any $F$-stable Borel subgroup of $\mathbf{G}$ is a Sylow $p$-subgroup of $\mathbf{G}^F$.

Steinberg's "Lecture Notes on Chevalley Groups" the Corollary of Lemma 54 on page 132. There is possibly a more modern reference.

EDIT: Sorry the reference to Steinberg is not sufficient as he does not treat arbitrary finite reductive groups. This follows from a theorem of Rosenlicht about the fixed points of a unipotent group under a Frobenius endomorphism. If $\mathbf{G}$ is a connected reductive algebraic group with Frobenius endomorphism $F : \mathbf{G} \to \mathbf{G}$ and $\mathbf{U} \leqslant \mathbf{G}$ is an $F$-stable unipotent subgroup then the finite group $\mathbf{U}^F$ has order $q^{\dim\mathbf{U}}$, (see Geck's book "An Introduction To Algebraic Geometry and Algebraic Groups" - Theorem 4.2.4). Now take $\mathbf{G} = \mathrm{GL}_n(\overline{\mathbb{F}}_p)$ and $F$ to be the map given by $F(x_{ij}) = (x_{ij}^q)$ where $q = p^a$ for some integer $a$ then $\mathbf{G}^F = \mathrm{GL}_n(q)$. The subgroup $\mathbf{B} \leqslant \mathbf{G}$ of upper triangular matrices is an $F$-stable Borel subgroup of $\mathbf{G}$ and its unipotent radical is the subgroup of all upper uni-triangular matrices. This has maximal dimension amongst all unipotent subgroups of $\mathbf{G}$, which can be seen in the following way. Assume $\mathbf{V} \leqslant \mathbf{G}$ is a unipotent subgroup of $\mathbf{G}$ then as $\mathbf{V}$ is solvable it is contained in a Borel subgroup $\mathbf{B}'$ of $\mathbf{G}$. As $\mathbf{V}$ is a unipotent subgroup of $\mathbf{B}'$ it is contained in the unipotent radical of $\mathbf{B}'$. As all Borel subgroups of $\mathbf{G}$ are conjugate we have the unipotent radicals of $\mathbf{B}$ and $\mathbf{B}'$ have the same dimension, hence we have $\dim\mathbf{V} \leqslant \dim \mathbf{U}$. Now applying Rosenlicht's theorem we have $\mathbf{U}^F$ is a Sylow $p$-subgroup of $\mathbf{G}^F$, which is precisely the subgroup in question. This approach has the advantage that it works for all finite reductive groups.

Steinberg's "Lecture Notes on Chevalley Groups" the Corollary of Lemma 54 on page 132. There is possibly a more modern reference.

EDIT: Sorry the reference to Steinberg is not sufficient as he does not treat arbitrary finite reductive groups. This follows from a theorem of Rosenlicht about the fixed points of a unipotent group under a Frobenius endomorphism. If $\mathbf{G}$ is a connected reductive algebraic group with Frobenius endomorphism $F : \mathbf{G} \to \mathbf{G}$ and $\mathbf{U} \leqslant \mathbf{G}$ is an $F$-stable unipotent subgroup then the finite group $\mathbf{U}^F$ has order $q^{\dim\mathbf{U}}$, (see Geck's book "An Introduction To Algebraic Geometry and Algebraic Groups" - Theorem 4.2.4). Now take $\mathbf{G} = \mathrm{GL}_n(\overline{\mathbb{F}}_p)$ and $F$ to be the map given by $F(x_{ij}) = (x_{ij}^q)$ where $q = p^a$ for some integer $a$ then $\mathbf{G}^F = \mathrm{GL}_n(q)$. The subgroup $\mathbf{B} \leqslant \mathbf{G}$ of upper triangular matrices is an $F$-stable Borel subgroup of $\mathbf{G}$ and its unipotent radical is the subgroup of all upper uni-triangular matrices. This has maximal dimension amongst all unipotent subgroups of $\mathbf{G}$, which can be seen in the following way. Assume $\mathbf{V} \leqslant \mathbf{G}$ is a unipotent subgroup of $\mathbf{G}$ then as $\mathbf{V}$ is solvable it is contained in a Borel subgroup $\mathbf{B}'$ of $\mathbf{G}$. As $\mathbf{V}$ is a unipotent subgroup of $\mathbf{B}'$ it is contained in the unipotent radical of $\mathbf{B}'$. As all Borel subgroups of $\mathbf{G}$ are conjugate we have the unipotent radicals of $\mathbf{B}$ and $\mathbf{B}'$ have the same dimension, hence we have $\dim\mathbf{V} \leqslant \dim \mathbf{U}$. Now applying Rosenlicht's theorem we see that the order of $\mathbf{U}^F$ is a power of $p$. We can now apply the order formula for finite reductive groups, (see Geck - Corollary 4.2.5), and easily deduce that this is the maximal power of $p$ dividing the order of $\mathbf{G}^F$. This approach has the advantage that it works for all finite reductive groups.

Steinberg's "Lecture Notes on Chevalley Groups" the Corollary of Lemma 54 on page 132. There is possibly a more modern reference.

EDIT: Sorry the reference to Steinberg is not sufficient as he does not treat arbitrary finite reductive groups. This follows from a theorem of Rosenlicht about the fixed points of a unipotent group under a Frobenius endomorphism. If $\mathbf{G}$ is a connected reductive algebraic group with Frobenius endomorphism $F : \mathbf{G} \to \mathbf{G}$ and $\mathbf{U} \leqslant \mathbf{G}$ is an $F$-stable unipotent subgroup then the finite group $\mathbf{U}^F$ has order $q^{\dim\mathbf{U}}$, (see Geck's book "An Introduction To Algebraic Geometry and Algebraic Groups" - Theorem 4.2.4). Now take $\mathbf{G} = \mathrm{GL}_n(\overline{\mathbb{F}}_p)$ and $F$ to be the map given by $F(x_{ij}) = (x_{ij}^q)$ where $q = p^a$ for some integer $a$ then $\mathbf{G}^F = \mathrm{GL}_n(q)$. The subgroup $\mathbf{B} \leqslant \mathbf{G}$ of upper triangular matrices is $F$-stable and its unipotent radical is the subgroup of all upper uni-triangular matrices. This has maximal dimension amongst all unipotent subgroups of $\mathbf{G}$ hence by Rosenlicht's theorem we have $\mathbf{U}^F$ is a Sylow $p$-subgroup of $\mathbf{G}^F$, which is precisely the subgroup in question. This approach has the advantage that it works for all finite reductive groups.

Steinberg's "Lecture Notes on Chevalley Groups" the Corollary of Lemma 54 on page 132. There is possibly a more modern reference.

EDIT: Sorry the reference to Steinberg is not sufficient as he does not treat arbitrary finite reductive groups. This follows from a theorem of Rosenlicht about the fixed points of a unipotent group under a Frobenius endomorphism. If $\mathbf{G}$ is a connected reductive algebraic group with Frobenius endomorphism $F : \mathbf{G} \to \mathbf{G}$ and $\mathbf{U} \leqslant \mathbf{G}$ is an $F$-stable unipotent subgroup then the finite group $\mathbf{U}^F$ has order $q^{\dim\mathbf{U}}$, (see Geck's book "An Introduction To Algebraic Geometry and Algebraic Groups" - Theorem 4.2.4). Now take $\mathbf{G} = \mathrm{GL}_n(\overline{\mathbb{F}}_p)$ and $F$ to be the map given by $F(x_{ij}) = (x_{ij}^q)$ where $q = p^a$ for some integer $a$ then $\mathbf{G}^F = \mathrm{GL}_n(q)$. The subgroup $\mathbf{B} \leqslant \mathbf{G}$ of upper triangular matrices is an $F$-stable Borel subgroup of $\mathbf{G}$ and its unipotent radical is the subgroup of all upper uni-triangular matrices. This has maximal dimension amongst all unipotent subgroups of $\mathbf{G}$, which can be seen in the following way. Assume $\mathbf{V} \leqslant \mathbf{G}$ is a unipotent subgroup of $\mathbf{G}$ then as $\mathbf{V}$ is solvable it is contained in a Borel subgroup $\mathbf{B}'$ of $\mathbf{G}$. As $\mathbf{V}$ is a unipotent subgroup of $\mathbf{B}'$ it is contained in the unipotent radical of $\mathbf{B}'$. As all Borel subgroups of $\mathbf{G}$ are conjugate we have the unipotent radicals of $\mathbf{B}$ and $\mathbf{B}'$ have the same dimension, hence we have $\dim\mathbf{V} \leqslant \dim \mathbf{U}$. Now applying Rosenlicht's theorem we have $\mathbf{U}^F$ is a Sylow $p$-subgroup of $\mathbf{G}^F$, which is precisely the subgroup in question. This approach has the advantage that it works for all finite reductive groups.