Because you've said that the input pin have it's own resistors.
That isn't exactly what I said. I said the input has an equivilent input impedance of 100 Mohms. If you use internal pullups they are 50k ohms.
With an input impedance of 100 Mohms, 5V connected directly to the pin would draw
I= V/R = 5V/100 Mohms = 0.00000005 A (50 nA)
Can you handle that ?
Assuming you haven't enabled the internal pull-up resistors? It will work, but not well. They have very high impedance as inputs but that also means they are very sensitive. So if you leave a pin "floating" (not connected to anything when the button isn't pressed) it will give off random values. Often messing with your program.
A floating pin will be read as a "0" (LOW).
An input should never be left floating. A pullup or pulldown resistor (10k or larger) should be used.