What is the minimal degree $ d$ required so a B – tree with $ 44*10^6$ keys will have a height $ h$ , such that $ h\leq 5$

My attempt was to build the tallest tree possible with minimum degree $ d$ and $ n = 44,000,000$ keys and then solve for $ d$ . That would mean any other tree with a minimal degree $ d’$ such that $ d’\geq d$ and $ n$ keys will be shorter than the one I built:

at depth 0 , we have the root and that’s $ 1$ node

at depth 1, we got exactly $ 2$ nodes

at depth 2, since we’re going for the tallest tree each node will have a minimal number of keys so $ d-1$ keys each, that means $ d$ children each so a total of $ 2d$ nodes.

at depth 3, following the same reasoning , $ 2d^2$ nodes.

…

at depth $ h$ , there are $ 2d^{h-1}$ nodes

total number of keys is :

$ n = 1+ (d-1)\sum_{k=0}^{h-1} {2d^k} = 1 + (d-1) \frac{2(d^h-1)}{d-1} = 2d^h-1 = 44*10^6 $

so:

$ 2d^5-1=44,000,000 $

$ d= 29.4 $

$ d\geq 30$

is that even correct ?