First note that natural logarithm is an incresing function (1) since for all
Now:
Thus the Natual Logarithm has the following property, for :
Thus: for any natural number (2)
Suppose we want to prove that by definition.
We have to prove that, for all there is a number such that if we have (3)
Setting by (1) and (2) we know that the condition (3) holds.
Just see that iff and since the function is increasing if then
So by definition it follows that (4)
For the second part set and use the property shown in the second line, so that and since we have and the assertion follows by (4)
The problem with this approach is that the typical proof of the divergence of the harmonic series uses the integral test which uses . So it is kind of circular. But not really because there are some proofs which show divergence of harmonic series without integral tests.
Good job. But there is just a minor problem here. What does mean? You are assuming the definition of exponents for arbitrary real powers. However, the generalized definition needs further properties of which you are trying to avoid in the first place.
Here is a general theorem. Let be a function (defined on ) such that (i) is increasing (ii) is not bounded from above. Then . Here is the proof. We prove this directly by definition. Let . Then since is not bounded from above there is such that . Then if it means . Proof complete. Now we can apply this to function. Since this function is increasing this satisfies (i). Now we need to satisfy (ii) i.e. the function is unbounded this takes a little bit more work. Notice that (when ) and . Thus, . Since and it means there is a number, call it , such that (by property of continous functions). Then we see as PaulRS showed that for all positive integers . But since we can make are large as we please it means is an unbounded function. The rest follows.