Proof by Induction for Natural Numbers Show that if the statement 
$$1 + 2 + 2^{2} + ... + 2^{n - 1} = 2^{n}$$
is assumed to be true for some $n,$ then it can be proved to be true for $n + 1.$ Is the statement true for all $n$?
Intuitively, then I don't think it holds for all $n.$
 A: If we assume that $1+2+\cdots+2^{n-1}=2^n$, we can easily prove that $1+2+\cdots+2^n=2^{n+1}$.  For
$$1+2+\cdots+2^n=(1+2+\cdots+2^{n-1})+2^n=2^n+2^n=2^{n+1}.$$
But of course it is not true that $1+2+\cdots+2^{n-1}=2^n$.  The base case $n=1$ does not hold. For then $1+2+\cdots+2^{n-1}=1\ne 2^1$.
The whole point of this exercise is that in order to prove a result by induction, we must do the induction step (which worked) and we must verify the base case (which failed).
In fact, $1+2+\cdots+2^{n-1}=2^n-1$.  The proof is straightforward.
A: So, let $n = 3 \in \mathbb{N}$. Then we have, $$2^0+2^1+2^2=1+2+4=7 \neq 2^3$$ 
A: This statement is false. According to this statement:
$1+2^{1}=2^{2} \\
3=4$
which is false.
A: Let premise $P(n)$ be $\sum\limits_{k=0}^{n-1} 2^k = 2^n$
If we assume $P(n)$, for some $n{\geq}1$ , then $\sum\limits_{k=0}^n 2^k = 2^n+2^n = 2^{n+1}$, which means $P(n+1)$
So we can indeed show: $\forall n{\geq}1:P(n)\to P(n+1)$
This is not enough!   In order to prove $\forall n{\geq}1:P(n)$ we need to show $P(1)\;\wedge\; \forall n{\geq}1:P(n)\to P(n+1)$.
That is to say, the iterative step alone is not sufficient, a proof by induction also requires demonstrating some base case.
In this case, since $P(1)\equiv  (2^0=2^1)$, we cannot prove it.

Indeed, rather we can prove:
$$\forall n{\geq}1: \sum_{k=0}^{n-1} 2^k= 2^n -1$$
