Saying that "4 divides [itex]5^n- 1[/itex] is NOT just a "reference" to [itex]\frac{5^n- 1}{4}[/itex]! It is the statement that [itex]\frac{5^n- 1}{4}[/itex] is an integer. That is, [itex]\frac{5^n- 1}{4}= k[/itex] for some integer k. Further the "n+1" form of the formula is not [itex]\frac{5^{n+1}- 1}{4}+ (n+1)[/itex]. I don't where you got that additional "(n+1)"! Replacing n by n+1 in [itex]\frac{5^{n}- 1}{4}[/itex] is just [itex]\frac{5^{n+1}- 1}{4}[/itex].
Now, of course, you want to "algebraically" go back to the "[itex]5^n[/itex]" and to do that use the fact that [itex]5^{n+1}= 5(5^n)[/itex].
It will be helpful to use [itex]5(5^n)- 1= 5(5^n)- 5+ 4[/itex].