Assume a linear scale, and assume the 1,000,000:1 range is between the highest luminance (1,000,000) to the lowest luminance (1) that isn't pure black.
(Pure black might be zero, but that's useless for computing a ratio, so it doesn't give you a resolution.)
Then you need 20 bits, because log_2 1e6 rounds up to 20 (equivalently, 2^20 <= 1,000,000 < 2^21).
If you used only 8 bits, the ratio between highest and lowest (non-zero) luminance would be only 256:1, or 1,000,000:3,906.
20 bits lets you represent more values between n and 1,000,000 times n than 8 bits does, but it still doesn't let you represent them all, because these are real numbers and there an infinite number of values between n and 1,000,000 times n.
20 bits lets you represent all the values which are integral in base 10.... but so what? Why is the fact that a value is integral in a particular base make it the minimum resolution difference that you need to represent? Why do you need to quantise at that level? There is a luminance of 1.5 times n, isn't there? But 20 bits doesn't let me represent that. 21 bits would. See what I'm saying?
I understand you, I had the same thoughts as you in the past.
The fact is that if you want to have good highlights (beautiful clouds, candles, sun glares etc..) in your image when shooting with a camera, you have to "expose by the right" i.e. if the scene is more dynamic => reduce the amount of light incoming into the camera => your shadows will be crushed down so less precise => you need better color resolution if you want to "save" your shadows OR you need to use another specific colorspace (S log shaped)
(Pure black might be zero, but that's useless for computing a ratio, so it doesn't give you a resolution.)
Then you need 20 bits, because log_2 1e6 rounds up to 20 (equivalently, 2^20 <= 1,000,000 < 2^21).
If you used only 8 bits, the ratio between highest and lowest (non-zero) luminance would be only 256:1, or 1,000,000:3,906.