Here is a basic geometric argument. Consider the sum 3^2 + 4^2 = 5^2. We can represent this as three "squares" of ones, such as:
1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 + 1 1 1 1 = 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1
Where each number is represented as the sum of ones in each square.
Now, consider the case of a^3 + b^3 = c^3. For illustrative purposes, let's assume that, for instance, 3^3 + 4^3 = 5^3. If this was true, it could be written in the form:
3 3 3 4 4 4 4 5 5 5 5 5
3 3 3 + 4 4 4 4 = 5 5 5 5 5
3 3 3 4 4 4 4 5 5 5 5 5
4 4 4 4 5 5 5 5 5
5 5 5 5 5
In essence, we have our squares again, only the "units" are now equal to the length of a side.
Assumption -- all powers > 2 can be expressed this way, meaning essentially as squares but with the appropriate units. So, for instance, the number 3^5 can be expressed as:
27 27 27 9 9 9 3 3 3 1 1 1
27 27 27 and 3^4 = 9 9 9 and 3^3 = 3 3 3 and 3^2 = 1 1 1
27 27 27 9 9 9 3 3 3 1 1 1
So, then, for each triple (a,b,c), this assumption implies that if a solution to a^x + b^x = c^x exists, then a^2 + b^2 = c^2 must also be true. Since both of those things cannot be true, this shows that triples of higher powers cannot exist.
What I don't know is if this is sufficient to show that every possible triple of a higher power therefore cannot exist.
EDIT: This is way too "easy", I think, to be a proof of Fermat's Last Theorem by contradiction. I feel like I'm making an error someplace, namely in the assumption that if (a,b,c) is a "triple" in some higher power, then it also must be a Pythagorean triple.
EDIT2: I have removed the argument that a "triple" in higher powers implies that we can also do a^2/a + b^2/b = c^2/c and arrive at a+b=c, hence a contradiction.
EDIT3: I have changed the title to better reflect my actual question.
If $a^n+b^n=c^n$, then $\sqrt[n]{a^n+b^n}=c$ : true
$\sqrt[n]{a^n+b^n}=\sqrt[n]{a^n}+\sqrt[n]{b^n}$ : false