You are not wrong. However voltage play huge role in the high power system, especially a DC system.

for the same 70kW, doubling voltage reducing the current by half.

1) Let's assume perfect power factor of 1.0 for simplicity. P = I^2 * R, high school physic. P = Square of Current multiplied by Resistant.

Use V=IR

70,000W with 400V, current = 175A
70,000W with 800V, current = 87.5A

2) Assume gauge 0 cable (that's very thick 0.855cm in diameter cable on each polarity), the resistance is about 0.0004 ohm per meter at room temperature 25C. Assume a 10m cable, that's 0.004 ohm, round trip = 0.008 ohm.

Use P = I^2 * R

400V, power loss over cable = 175^2 * 0.008 = 245W
800V, power loss over cable = 87.5^2 * 0.008 = 61.25W

You see, significantly less loss over the charging cable wires with 800V.
Furthermore the loss power will dissipate as heat, cause the wire to heat up and increase the wire resistance -> even more loss -> even more heat.

3) Let's not forget the voltage drop over the cable. In the school, on electronics, the current is so small and insignificant, we always assume wire resistance = 0, voltage drop over the wire = 0V.

In high power system, it is different story.

Use V = IR

400V voltage drop over the wire = 175 * 0.008 = 1.4V. Voltage at load end = 398.6V
800V voltage drop over the wire = 87.5 * 0.008 = 0.7V. Voltage at load end = 799.3V

You see wire itself cause voltage drop on the high power system, thus we always have something called remote sensing, where we have high impedance load that run in parallel as the actual application load. Such the the current over the high impedance load is small and negligible, then we can measure the load end voltage drop, and compensate it by increase the source voltage.

So source run slightly higher voltage at source to allow load to actually get 400V, 800V.

4) Lastly, running at 800V system also allow thinner wire due to lower current on the car itself. Thus lower loss + lighter car.

5) So, 70kW at 800V will charge faster than 70kW at 400V due to higher efficiency (provided the the bottleneck is not at battery side, and assume same battery thermal management).

50k for 1 whole block of battery? My understanding is that most batteries are now modular. Example, Volvo's PHEV, to replace all is about 40+k, but per block/module is about 5k, which i think still kinda reasonable, unless you unfortunate at the max level.