Fake Proof: $a^n = 1$ f or all nonnegative integers $n$ 
Find the flaw with the following "proof" that $a^n = 1$ for all nonnegative integers $n$, whenever $a$ is a nonzero real number.
Basis Step: $a^0 = 1$ is true by definitino of $a^0$.
Inductive Step: Assume that $a^j = 1$ for all nonnegative integers $j$ with $j \leq k$. Then note that
  $$
  a^{k+1} = \frac{a^k \cdot a^k}{a^{k-1}} = \frac{1 \cdot 1}{1} = 1
$$

Where is the flaw in the above proof? My answer is that the basis step needs to iterate nonnegative integers beyond $0$, like $1$, $2$, etc. 
Any thoughts?
 A: $a^1=\cfrac {a^0\cdot a^0}{a^{-1}}$ and your hypothesis doesn't cover the case of $a^{-1}$. 
A: The inductive step requires $k\geq1$, because of the denominator (you are using that $a^{-1}=1$, which forces $a=1$). But you want to start your induction at $k=0$. So the base step and the inductive step are not connected. 
A: Clearly the induction fails on the first step, since when you only have $k = 0$, you cannot make use of $a^{k-1} = 1$.
A: 2 things wrong:
1)You can not assume anything more in your induction step than was shown in the base step.
In here you are assuming that k-1 is also nonnegative.  That is not the case for k = 0 so we can not assume that.
2) Even more basic.  When you assume a base step is true for k, you can not assume you have shown it is true for all j <= k.  You can only assume it is true for k and only k.  This is important if the inductive step requires not just k but k- 1, then your base step must not merely show for k = 1 (or zero) but for k = 0 (or negative 1) as well.  Or equivalently for k=1 and k = 2.
If you do show for both k =1 and k=2, your induction is sound.
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This reminds me of the proof that all the marbles in a bag of marbles are the same color.
Base case: n = 1. All the marbles in a bag  of just one marble are same color.  (Because there is only 1 marble so there can't be more than 1 color.)
Inductive case:  Assume all the marbles in a bag of n marbles are the same color.  Take a bag of n + 1 marble.  Take a marble out.  You now have a bag of n marbles.  Therefore they are all the same color.  Remove one of these marbles.  Put the original marble you removed back in.  You have a bag of n marbles so they are all the same color.  No return the 2nd marble.  It was the same color so now all the marbles in the bag of n+1 marbles are the same color.
Therefore by induction all the marbles in any size bag of marbles are the same color.
