I have seen two different definition of $C^{k,\gamma}(\Omega)$:
semi-norm $$[u]_{C^{0,\gamma}}:=\sup_{x,y\in \Omega,x\neq y}\frac{|u(x)-u(y)|}{|x-y|^\gamma}$$
one defines $$\|u\|_{C^{k,\gamma}(\Omega)}:=\max_{1\le|\alpha|\le k}\sup_{x\in\Omega}\|D^\alpha u(x)\|+\max_{1\le|\alpha|\le k}[D^\alpha u]_{C^{0,\gamma}}$$
another defines: $$\|u\|_{C^{k,\gamma}(\Omega)}:=\max_{1\le|\alpha|\le k}\sup_{x\in\Omega}\|D^\alpha u(x)\|+\max_{|\alpha|= k}[D^\alpha u]_{C^{0,\gamma}}$$
are they essentially equivalent?
Yes. (Assuming $\Omega$ is open. I would not comment on other cases).
It is quite obvious that $||u||_{C^{k,\gamma}(\Omega)}$ from the second definition is bounded from above (with a constant $C=1$) by the one from the first definition.
To see that the reverse inequlity is also true, is is sufficient to bound, for any $0\le l < k$,
$\max\limits_{|\alpha|=l}[D^\alpha u]_{C^{k,\gamma}}$ by $||u||_{C^{k,\gamma}(\Omega)}$ taken from the second definition.
Note this is actually a local question, i.e. a question about $x, y$ being close together.
To simplify notation fix some multi index $\alpha$ of length $l <k$ and let $\phi(x):=\partial_{x_1}^{\alpha_1}\cdots\partial_{x_n}^{\alpha_n}u(x) $.
Now fix some $x_0$ in $\Omega$ in order to estimate the term $$(*) = \frac{|\phi(x)-\phi(y)|}{|x-y|^\gamma}$$ in a neighbourhood of $x_0$.
One might be tempted to argue that this term can be estimated like
$$ \le \frac{|d\phi(x)(x-y) + o(|x-y|)|}{|x-y|^\gamma}\le |\phi|_{C^{l+1}(\Omega)}|x-y|^{1-\gamma}+ s_x(x-y) $$
The problem I'm having with this is the dependency of the remainder term $s_x$ on $x$. Instead you can apply the mean value theorem. Since $\Omega$ is open there exists an open convex neighbourhood (a ball) $U\subset\Omega$ of $x_0$.
In $U$ $$(*) = \frac{|\int_0^1\langle\nabla\phi(tx+1-t)y, x-y\rangle \,dt|}{|x-y|^\gamma}\le |\phi|_{C^{l+1}(\Omega)}|x-y|^{1-\gamma}$$ which can be clearly bounded by the norm taken from the second definition.