Can you use an input pin instead of ground in Arduino circuitry?

Beginner here: Would the circuit in the picture work or do input pins not absorb the energy as ground does? I know setting the pin as an input makes the direct 5V connection safe but maybe I'm mistaken. I'm trying to understand the nature of how these connections are made. Thank you in advance!

An input pin can only sense the voltage that is applied to it - whether it is above or below a pair of thresholds.

In your existing circuit when the button is not pressed the voltage sensed by the pin is 0V because it is connected to 0V (also known as Ground) through the resistor. When you press the button the input pin is connected directly to +5V, so the pin senses 5V.

If you remove the resistor or the connection to ground the pin can still sense 5V on it when the button connects it to 5V, but when you release the button the input pin then isn't connected to anything at all. It's not connected to 0V, so it can't sense what voltage is there - there simply isn't any voltage connected to it (no voltage isn't the same as 0V).

In this case the pin is said to be "floating". It's neither HIGH (5V) nor LOW (0V), it's just dangling around not knowing what it's doing.

If you want to get more technical and look at the mechanics of it, it's helpful to know that the gate of a MOSFET (which is what the input pin actually is) looks to all intents and purposes like a small capacitor. When you connect the input pin to +5V the capacitor is charged up (current flows into it from 5V to charge it). When you connect the input pin to 0V it discharges (current flows out of the capacitor down to 0V). When charged you have a HIGH and when discharged you have a LOW.

However when a charged capacitor has no connection it (for a certain amount of time, due to self discharge) keeps hold of its charge. So if you charge it with +5V and then disconnect it you will still get +5V on the capacitor. That is until it discharges through internal leakage currents.

Those leakage currents though are so small and the capacitor is so sensitive that the input pin acts like a radio antenna. It picks up noise from the surrounding area (electromagnetic waves) which can charge and discharge the capacitor pretty much at random. So at times you may see the input as HIGH and at other times as LOW, purely depending on chance. And the longer your wires the more noise it will pick up.

The worst source of noise is the 50/60Hz mains hum that is all around you. You can see this in reality if you graph the voltage on a disconnected ADC pin and bring your hand close to it - you act like a capacitively coupled antenna and you can see the mains waveform on your graph.

It's the same on a digital input pin, except that it converts it to a square wave as it passes the thresholds (you might like to look at schmitt triggers for more detail about how it does that).

So adding the pull down (or pull up if you prefer) resistor is critical to making the input pin see what you want it to see, and not just random noise.

Since the input is like a capacitor selecting the right value of resistor can be quite important in some situations. For a simple button like this there's nothing to worry about, but for something like a rotary encoder where the input signal may change more rapidly the resistor value does become more of a concern. Have too high a value and the gate will take too long to discharge (or charge if it's a pull up resistor). Too low and too much current could be wasted through the resistor when the button (or encoder) is active (potentially damaging the button if it's a low current sensitive device).

If you remove the ground wire, digital pin 2 will be left "floating" until the button is pressed, and this is not a good thing to do: