I've got the first part of the proof.
So we know that the odd exponents will give us an irrational answer, and the even ones will give us a rational answer. Therefore we have an alternating sum of rationals and irrationals. I can prove that an irrational plus a rational is irrational, so now we can combine all the alternating sum into a sum of irrationals. I can't prove that for this specific case that an irrational plus an irrational equals an irrational.
On
Tricks, tricks, tricks.Here, it's the geometric summation. $$ \sqrt{2} + \sqrt{2^2} + \ldots \sqrt{2^n} = 2^{\frac 12} + 2^\frac{2}{2} + \ldots + 2^\frac{n}{2} $$ Use the geometric series summation formula: $$ 2^{\frac 12} + 2^\frac{2}{2} + \ldots + 2^\frac{n}{2} = (2^\frac 12)\frac{2^\frac{n+1}{2} - 1}{2^{\frac 12} - 1} $$ Now, rationalize by multiplying top and bottom by $2^{\frac 12} + 1$: $$ (2^\frac 12)\frac{2^\frac{n+1}{2} - 1}{2^{\frac 12} - 1} \times \frac{2^{\frac 12} + 1}{2^{\frac 12} + 1} = 2^{\frac 12} (2^{\frac 12} + 1)(2^{\frac{n+1}{2} - 1}) $$ expand this to get: $$ {\frac 12} (2^{\frac 12} + 1)(2^{\frac{n+1}{2} - 1}) = 2^{n/2}(1 + \sqrt 2) = 2^{\frac n2} + 2^\frac{n+1}{2} $$
Now, suppose that $n$ is even: Then $2^{n/2}$ is an integer, so the answer is of the form $a+b \sqrt 2$ where $a,b$ are integers. If this were rational, then $\sqrt 2$ would also be rational. Contradiction.
Now, suppose that $n$ is odd: Then $2^\frac{n+1}{2}$ is an integer, so again the answer is of the form $a+b \sqrt 2$ where $a,b$ are integers. If this were rational, then $\sqrt 2$ would also be rational. Contradiction yet again.
Hence, we can conclude that for all $n$, the quantity given is irrational.
The sum can be expressed as $L+M\times\sqrt2$, where $L$ and $M$ are integers. If $L+M\times\sqrt2$ is rational, then $\sqrt2$ has to be rational. But $\sqrt2$ is not rational, thus the sum is not rational. I assume you can use the fact $\sqrt2$ is irrational.