This is a great question. Let's show this in a really straightforward way that I find intuitive. We want to show that there are many n that are very near multiples of pi (in fact, arbitrarily close).
Start with 3. We know that 3 is pretty close to pi: it's off by about 0.14. This means 6 will be off by about 0.28, 9 by about 0.42, and k * 3 will be off by about k * 0.14... almost. We can't be off by more than pi, since we're just talking about multiples of pi. [In fact, we can't be off by more than pi/2, since we'll always be within pi/2 of a multiple of pi (within pi/2 below, and within pi/2 above). But we ignore this right now]
We take advantage of this looping behavior. We start off with error about 0.14. Let's go up a loop. Notice that 22 * 0.14 = 3.08 < pi < 23 * 0.14, so that some multiple of pi is closer to 22 * 3 and 23 * 3 than any multiple of pi is to 3. In particular, 22 * 3 is within about 0.026 of 21 * pi. Notice also that 22 is the closest integer to pi/(pi - 3). This is not an accident (do you see why?), and we can repeat.
Now we know that 22 * 3 = 66 is within 0.026 of a multiple of pi (namely, 21 * pi). Notice that pi/(22 * 3 - 21 * pi) is about 118.06, so that 118 is the nearest integer. Using == to mean approximately, pi/(22 * 3 - 21 * pi) == 118, rearranging into pi == (22 * 3 - 21 * pi) * 118, rearranging into (118 * 22) * 3 - (118 * 21 + 1) * pi == 0 (in fact, it's really really small).
So now we see that 118 * 22 * 3 is much closer to a multiple of pi. Each time we repeat, we'll cut the distance to pi by a factor of at least 2, so this rapidly converges to multiples of pi.
Since this clearly continues and we get closer and closer to multiples of pi very rapidly, tan(n) is unbounded.
As an aside
These numbers are sort of special. The first approximation was 3. Not so special. We then said that 22 * 3 == 21 * pi, or rather that 22 * 3 / 21 = 22/7 == pi. This is a well-known approximation, and is sometimes mistaught as actually being pi.
The next one was 118 * 22 * 3 / (118 * 21 + 1) == pi, or rather that 7788/2479 == pi.
In these forms, these look like convergents of continued fractions approaching pi. In fact, since we allow the sign to alternate, these are better (but less well-behaved) than the classical continued fraction for pi.
In a different form, inverting