Why does $2+{{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}=(1+1)^n$ $${{n}\choose{0}}+ {{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}+{{n}\choose{n}}=(1+1)^n$$


I don't see why this is true, because (if I'm not mistaken)

\begin{align}&amp;{{n}\choose{0}}+ {{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}+{{n}\choose{n}}\\&amp;=1+{{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}+1\\
&amp;=2+{{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}
\end{align}

So my question is why

$$2+{{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}=(1+1)^n\;\;\;\;\text{(1)}$$

If anyone requires any context, I am reviewing some set theory, and I came across the proof that the power set of a set $S$ with $n$ elements  had $2^{n}$ elements. Just to reiterate, I'm not looking for a proof of the cardinality of the power set. I just want to know, algebraically, why $(1)$ is true.

Thanks.
There is an easy way to prove the formula.

Suppose you want to buy a burger and you have a choice of 6 different condiments for it - mustard, mayonnaise, lettuce, tomatoes, pickles, and cheese. How many ways can you choose a combination of these condiments for your burger?

Of course, you can choose either 6 different condiments, 5 different condiments, 4 different condiments, etc. So the obvious way to solve the problem is:

$\begin{align}{{6}\choose{6}} + {{6}\choose{5}} + {{6}\choose{4}} + {{6}\choose{3}} + {{6}\choose{2}} + {{6}\choose{1}} = \boxed{64}\end{align}$

But there is a better way. Imagine 6 spaces for 6 condiments:

_____ _____ _____ _____ _____ _____

For every space, there are $2$ possible outcomes: Yes or No, meaning the condiment was chosen or the condiment was not chosen. With $2$ possible outcomes for each space, there are $2^6 = \boxed{64}$ possible ways.

We know that both ways have foolproof logic and will both give identical answers no matter how many condiments there are. So this means we have proven:

$\begin{align}\sum_{k = 0}^{n} \binom{n}{k} = 2^n\end{align}$
Well it just follows from the general formula:

$(a+b)^n=\sum\limits_{k=0}^n \binom {n} {k}\,a^k\,b^{n-k}$,

which you can prove by induction see  here .