Can we find every finite sequence of $\pi$ within $\pi$? Inspired by one of the top posts I was wondering if $\pi = 3.14159\dots$ were normal--as in you could find every finite string of numbers within $\pi$'s digits--would that mean we could find every finite sequence of the digits of $\pi$ in $\pi$'s digits. Would we find 
$$3141592653...\left(\text{$n$-th digit of $\pi$} \right)$$  
For any $n$? Also is there an example of a number that is normal that is easy to prove to be normal? I'm unfamiliar with the property.
 A: The answer to your question is yes: the first billion digits of $\pi$ forms a finite block of base-$10$ digits, and every finite block of base-$10$ digits must occur infinitely often in any normal-to-base-$10$ number. (I wonder if you asked exactly what you wanted to?)
A quick comment at this point: it looks from your question like you've gotten the definition of "normal" incorrectly. Normality and its relatives don't just assert that lots of variety occurs, they assert that it occurs statistically: e.g. simple normality to base $10$ implies that asymptotically $1\over 10$th of the number's digits in base $10$ are $2$, not just that infinitely many digits are $2$. Precisely:


*

*A number is simply normal to base $b$ if each digit in base $b$ occupies $1\over b$th of the places of the base-$b$ expansion of the number (asymptotically speaking). Note that the number whose binary expansion is $0.01001000100001..._2$ is not simply normal to base $2$, even though both $0$ and $1$ occur infinitely often in its expansion, since their distribution isn't right: the asymptotic probability of a digit being $0$ is $1$, not ${1\over 2}$.

*A number is normal to base $b$ if it is simply normal to base $b^n$ for all $n$. Intuitively, this means that not only are the individual base-$b$ digits stochastically distributed, but so are the finite blocks of digits; e.g. $1\over 3$ is simply normal to base $2$ but not normal to base $2$ (since e.g. the block "$00$" doesn't appear anywhere in the binary expansion of $1\over 3$, let alone the required "$1\over 4$th of the time").

*A number is normal (or absolutely normal) if it is simply normal to all bases.

Now let me address your secondary question:

Also is there an example of a number that is normal that is easy to prove to be normal? 

Despite their objective commonality (in a precise sense, "most" real numbers are normal), verified examples of naturally-occurring normal numbers are hard to come by. Champernowne's constant is easy to prove "normal to base $10$," but isn't exactly "naturally occurring" and it is not known to be normal to all bases. Examples of absolutely normal numbers can be explicitly constructed; however, these examples are even less naturally-occurring than Champernowne's constant.
Meanwhile, it's conjectured that all irrational algebraic numbers are absolutely normal, but at the same time there is no known irrational algebraic absolutely normal number.
In general, these slides (and the sources they mention) are worth looking at. It's all a very intriguing situation, where we have heuristic evidence suggesting that a truly vast class of naturally-occurring numbers are absolutely normal but we currently lack the tools to verify any of those cases, and even the construction of explicit non-natural examples faces deep mathematical problems/questions.
