The physical size of a binary LUT is exponential in its number of inputs. In particular, every time you add another input, the size doubles. To go from 10 inputs to 20, the size of each LUT would go up by a factor of 1024, and you'd barely be able to fit any of them onto the FPGA anymore.
You don't typically see LUTs with more than 6 inputs on FPGAs, as this already requires 64 bits of storage per LUT. It's much more efficient to compose circuits from multiple LUTs than to put everything into a single huge LUT.
As a simple example, let's say that you want to build a 16-input AND gate on an FPGA.
If the FPGA has 16-input LUTs, you can do it in a single LUT, which would contain 2^16 = 65536 SRAM cells.
If, however, the FPGA only has 4-input LUTs, you'll need to build a LUT cascade. Each LUT, when configured as an AND gate, compresses 4 inputs down to 1 output. To get from our total of 16 inputs down to 1 output, we need two layers of compression. This requires 5 LUTs in total. Each 4-input LUT consists of 2^4 = 16 SRAM cells, so in total, our 16-input AND gate requires 5*16 = 80 SRAM cells in this case.
By shrinking our LUT size from 16 inputs to 4, we were able to shrink the silicon area used by our circuit from 65536 SRAM cells down to just 80, even though we used more of those smaller LUTs.
Additionally, FPGAs have mechanisms to combine multiple smaller LUTs into larger ones if a circuit really requires big LUTs (for example to store constant data). Modern Xilinx FPGAs have 5-input LUTs, but can internally connect multiple of them together to form LUTs with up to 8 inputs within a single logic slice (at least if I remember this correctly). Bigger ones can be created across multiple slices, too, by using the FPGA's routing fabric.