The answer lies in **Kirchhoff's Current Law** (KCL), from which we can state that within a closed circuit, the current is *equal at each point in the circuit*. The current that flows in is equal to the current that flows out.

Yes, kind of like water into and out of a garden hose. Push current in, the same current pushes out. What will differ is the *pressure* at each end. Back in the electrical world, voltage is *pressure*, while current is *flow*.

Consider an LED in series with resistor, connected to a battery. That's a closed circuit. KCL tells us that LED and resistor will see the same current, *regardless of connection order*.

Now, each element has its own voltage drop.

  * LED voltage drop is its forward threshold voltage (Vf)
  * Resistor voltage drop is proportional to current (E = I*R).

The total voltage drop for the two will be Vf(LED) + I*R. 

Notice that '+' sign. You may recall that addition is **commutative**. It doesn't matter what order you do the addition. You get the *same total voltage drop*.

Further, regardless of connection order, because we know based on KCL that the currents will be the same, so will the terminal-to-terminal voltage drop of each. Vf will not change, nor will E = I*R. The only difference will be the voltage seen between the resistor and LED relative to the power supply.

Here's a quick sim to show what's up ([simulate it here][1]):

[![enter image description here][2]][2]

You'll notice in each case that:

   * LED forward drop is the same
   * Resistor IR drop is the same
   * The currents are equal everywhere

The only difference we see is the voltage between the LED and resistor, since that shifts with the component order.

  [1]: https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgxCVxju8qwvcCjwhuAEwYAzAIYBXADYAXJgobjwULTEjsATj24oUNQ%20AyEtYeGwBuIYpZNVH4MNypUaVJJ6jQENgB3B0tWbldCU0hgsyizfihYwSMTM2ck7AwXS3iUkDztNiyBIXdRfLALTU9i7PTvULFGiFqAcyaM1xQES1qABzEMPipMbnKQelqQ-MSxkESYmbKRIfHVpfS0%20eNo5LLq%20aq%20tkGjw%20GxNKmkwd6qRPvhURvNmkTHwgfp2kSJp4mMQ62DwpgyILBvSKJV%20o0aEOaAmh9XeAlWqPMYWRVAR8QxhR%20GN2sMRSRCCPCk1BBT29gR4OplL8jV82gCbAMGImGMWyDgHC4JOOJOJEmk8mUqnUNTZulOFC%20mIVo2uRRCXOqAOqbzmaSexJ1VDxH1qQA
  [2]: https://i.sstatic.net/T1WM2.png