Question: I know the answer to this question is $1-(35/36)^n$. This is the concept behind it:

> 
> Probability of not getting a double six on a roll of two dice = 1–1/36 = 35/36.
> Therefore the probability of not getting a double six on n rolls =$(35/36)^n$
> Therefore the probability of getting at least a double six on n rolls =$1 - (35/36)^n$
> 
> 
> 

Now I have this dumb doubt: If for one throw of the pair of dice, the probability of getting a double six is $1/36$, then why is it not $1/36^n$ for n throws by the same logic as above?

Answer: 
Here, $v\_1, v\_2, \ldots, v\_n$ are $n$ vectors.

Let us fix some notation first.

We define $T = \mbox{span}\left\{ v\_1, v\_2, \ldots, v\_n \right\}$.

It is easy to verify that $T$ is a subspace of $V$.

The question is this: How to show that $T$ is the smallest subspace of $V$ containing the $n$ vectors $v\_1, v\_2, \ldots, v\_n$ ?

Indeed, let us take any subspace $S$ of $V$ that contains the $n$ vectors $v\_1, v\_2, \ldots, v\_n$.

Since $S$ is a subspace of $V$ and $v\_1, v\_2, \ldots, v\_n \in S$, $S$ must contain all the linear combinations of the $n$ vectors $v\_1, v\_2, \ldots, v\_n$.

In other words, for any set of $n$ scalars $\alpha\_1, \alpha\_2, \ldots,
\alpha\_n$,

$$
v = \alpha\_1 v\_1 + \alpha\_2 v\_2 + \cdots + \alpha\_n v\_n \in S.
$$

This essentially means that $T \subset S$, since $v$ is a general element in the subspace $T$.

Since we have shown that $T$ is a subset of any subspace $S$ of $V$
containing the vectors $v\_1, v\_2, \ldots, v\_n$, we have established that $T = \mbox{span}\{ v\_1, v\_2, \ldots, v\_n \}$ is the smallest subspace of $V$ containing the vectors $v\_1, v\_2, \ldots, v\_n$. $ \ \ \ \ \ $$\blacksquare$
The confusion you are having is that you are checking the equivalence of one definition with itself.

If you define $span\{v\_{1},v\_{2},...,v\_{n}\}$ as the smallest subspace containing $v\_{1},...v\_{n}$ then by definition it is the intersection of all subspaces containing $v\_{1},...v\_{n}$. You already say that you can prove that arbitrary intersection of subspaces is a subspace and hence it follows by definition that any subspace containing $v\_{i}$ is the intersection. Hence $\displaystyle\text{span}\{v\_{1},v\_{2},...,v\_{n}\}=\bigcap\{\text{S is a subsapce of V containing}\,v\_{1},...v\_{n}\}$.

Now if you define the span as the set of all linear combination of $\{v\_{i}\}$ then you can show that this set is actually the smallest subspace containing $\{v\_{1},v\_{2},...,v\_{n}\}$ and hence equals the intersection.

That is you need to check the equivalence of two definitions :-

1. $\displaystyle\text{span}\{v\_{1},..,v\_{n}\}=\{\sum\_{i=1}^{n}c\_{i}v\_{i}\,\,,c\_{i}\in\mathbb{F}\}$

2.$span\{v\_{1},...v\_{n}\}$ is the is the smallest subspace containing $v\_{1},...v\_{n}$.

Infact you can define this for arbitrary collection of vectors $\{v\_{i}\}\_{i\in I}$ where $I$ is some arbirary indexing set.

Then you can again check the equivalence of two definitions:-

1. $\displaystyle\text{span}\{v\_{i}\}\_{i\in I}=\{\sum\_{i\in I}c\_{i}v\_{i}\,\,,c\_{i}\in\mathbb{F},c\_{i}=0\,\text{for all but finitely many}\,i\in I\}=\{\sum\_{\text{finite}}c\_{i}v\_{i}\,\,,c\_{i}\in\mathbb{F}\}$.
2. $span\{v\_{i}\}\_{i\in I}$ is the smallest subspace of containing $\{v\_{i}\}\_{i\in I}$ which is equal to $\displaystyle\bigcap\{\text{S is a subsapce of V containing}\,\{v\_{i}\}\_{i\in I}\}$

So you see that if you prove $1\implies 2$ then it solves your "existence" crisis. And if you prove $2\implies 1$ then it means that if such a space exists then it is unique and is given by $1$.