Demonstration of sum of powers of $2$ Theorem : For every natural number $p$:
$$\sum^p_{i=0} 2^i = 2^{p+1}-1$$
I trieed to demonstrate the theorem using induction
Demonstration :
$1)$ If we have $p=0$ then we get $2^0=2^{0+1}-1$ that is always true.
$2)$ Supposing that the first statement is true then we get that
$$\sum^{p+1}_{i=0} 2^i = 2^{p+2}-1$$
Now we know that:
$$\sum^{p+1}_{i=0} 2^i - \sum^p_{i=0} 2^i = 2^{p+1}$$
and if that is true, must be also true that:
$$2^{p+2}-1-(2^{p+1}-1) = 2^{p+1}$$
$$2^{p+2}-1-(2^{p+1}-1)=2^{p+2}-2^{p+1}=2^{p+1}(2^1-2^0)=2^{p+1}$$
QED
Is this a valid demonstration?
 A: It looks like you are on the right track, but there are a few things to point out. This may be a matter of preference, but I think the induction step should be stated more clearly. Something like 

Induction Step: Suppose $\sum^{p}_{i=0} 2^i=2^{p+1}-1$ holds for all $p \in \{1,2,\ldots, k\}$. Then...

And proceed from there. At this point, I'm not sure of the benefit of saying $$\sum^{p+1}_{i=0} 2^i - \sum^p_{i=0} 2^i = 2^{p+1} \implies 2^{p+2}-1-(2^{p+1}-1) = 2^{p+1}$$
It appears to me that the above does not use the induction hypothesis, and any conclusion you reach seems a bit circular. What might be a better move is to write $$\sum^{k+1}_{i=0} 2^i= \sum^{k}_{i=0} 2^i+2^{k+1}$$ This is nice because you can say something about the quantity $\sum^{k}_{i=0} 2^i$ using the induction hypothesis. Can you take it from here?
A: Inductive hypothesis: $\quad\displaystyle\sum_{i=0}^p 2^i=2^{p+1}-1$ for some $p\ge 0$.
We must deduce that $\quad\displaystyle\sum_{i=0}^{p+1} 2^i=2^{p+2}-1$.
Decompose the sum: \begin{align*}\sum_{i=0}^{p+1} 2^i=2^{p+2}-1 &= \sum_{i=0}^p 2^i+2^{p+1}\\
&=2^{p+1}-1 +2^{p+1}\qquad\scriptstyle(\text{inductive hypothesis)}\\
&=2\cdot2^{p+1}-1=2^{p+2}-1.
\end{align*}
A: One can prove 
$$ \sum_{i=1}^{p}2^i=2^{p+1}-1 $$
without induction. Actually the same proof works for any geometric progression $a, aq, aq^2,\ldots$ if $q\ne 1$. For instance, we want to calculate the sum of the first $p+1$ terms.
Let
$$ a+aq+aq^2+\cdots+ aq^{p}=X. \tag1 $$
Multiplying (1) with $q$ and adding $a$ on both sides we get
$$a+q(a+aq+aq^2+\cdots+aq^p)=qX+a.$$
This gives
$$a+aq+aq^2+\cdots+aq^p+aq^{p+1}=qX+a,$$
i.e.,
$$ X+aq^{p+1}=qX+a.$$
It follows that
$$ a(q^{p+1}-1)=(q-1)X$$
and consequently
$$X=a\frac{q^{p+1}-1}{q-1}. $$
In the above case we have $a=1$ and $q=2$:
$$\sum_{i=1}^{p}2^i=X=2^{p+1}-1. $$
A: I think your first statement in part 2 puts the cart before the horse (you've assumed what you want to prove, or appear to).
The inductive step assumes that it holds for $p$:
$$\sum^p_{i=0} 2^i = 2^{p+1}-1$$
Now, given that, look at one side or the other for $p+1$.  I'll look at the left side:
$$\sum^{p+1}_{i=0} 2^i = \left(\sum^p_{i=0} 2^i\right) + 2^{p+1}$$
This doesn't assume what is to be proved.  I'm just splitting out the last term of the sum.
Now, substitute the assumed statement for $p$ on the right side:
$$\sum^{p+1}_{i=0} 2^i = 2^{p+1}-1 + 2^{p+1}$$
and simplify:
$$\sum^{p+1}_{i=0} 2^i = 2 \cdot 2^{p+1}-1 = 2^{p+2}-1 = 2^{(p+1)+1} - 1,$$
and you're done.
A: Another way to look at it is
$$\sum^{p+1}_{i=0} 2^i = 2 \sum^p_{i=0} 2^i +1$$
which can be used to go from $p$ to $p+1$.
