I found a formula for the height of a $k$-ary tree in a book "An Introduction to Data Structures and Algorithms" by J. A. Storer.
It computes the height as follows (page 224, chapter 7):
$$ \begin{align} h &= \log_k(L(n)) \\ &= \log_k\left(\frac{(k - 1) n + 1}{k}\right) \\ &= \log_k\left((k - 1) n + 1\right) - 1 \\ &= \left(\log_k(k-1) + \log_k(n)\right) - 1\\ &= (1 + \lfloor \log_k(n) \rfloor) - 1 \\ &= \lfloor \log_k(n) \rfloor \end{align} $$
Here $L(n)$ is the number of leaves in a complete $k$-ary tree of $n$ vertices.
Unfortunately I can't see the logic between steps 3 and 4: $$ \log_k\left((k - 1) n + 1\right) - 1 = \log_k((k-1) n) - 1 $$
Essentially it claims that $$\log_k(x + 1) = \log_k (x)$$
What do I miss here?
After some thought I still think it is a mistake to drop $+1$, but the final result should be correct, if we do following:
$$ \begin{align} \log_k\left((k - 1) n + 1\right) - 1 &= \log_k\left((k-1)\cdot n \cdot (1 + \frac{1}{(k-1)n})\right) - 1\\ &= \log_k(k-1) + \log_k{n} + \log_k (1 + \frac{1}{(k-1)n}) - 1 \\ &= \log_k\left((k-1)\cdot \left(1 + \frac{1}{(k-1) n}\right)\right) + \log_k n - 1\\ &= \log_k\left(k - 1 + \frac{1}{n}\right) + \log_k n - 1\\ &= \lfloor \log_k(n) \rfloor \end{align} $$
The last step is valid, because we know that
$$ 0 < k - 1 + 1/n \leq k $$