Computing modular exponent given order
There's a small chance you might be able to get a speed-up by using properties of arithmetic modulo $n^2$, depending on how $g$ was chosen.

Suppose $g=a(1+bn)$.  Then note that, by the binomial theorem,

$$g^x = a^x (1+bn)^x = a^x \sum_i {x \choose i} (bn)^i  = a^x (1 + bx n + n^2 \times \text{stuff})$$

so we see that

$$g^x = a^x (1 + bx n) \pmod{n^2}$$

Thus if you can compute $a^x \bmod n^2$ efficiently, you can compute $g^x \bmod n^2$ with just a tiny bit more work.  Here you want to take $x=mn$.  The simplification is that now $g$ can always be expressed in the form $g=a(1+bn)$ where $0 \le a &lt; n$.

What can we say about $a$?  Well, for one thing we know that $g^{kn}=1 \pmod{n^2}$, so it follows that $a^{kn}=1 \pmod{n^2}$.  Thus, the order of $a$ modulo $n^2$ has to be a divisor of $kn$.

If the order of $a$ is exactly $kn$, we haven't gotten anywhere.

But if the order of $a$ is a strict divisor of $kn$, we can get some speedup.  We first reduce $mn$ modulo the order of $a$, then compute $a^{mn} \bmod n^2$, and finally use the relationship above to get the value of $g^{mn} \bmod n^2$.  (For instance, if you get lucky, it is possible that the order of $a$ might be $k$, and then you'll get a non-trivial speedup.)

Whether this yields a speedup in practice will depend on the value of $g$ and how $g$ was chosen; since that wasn't specified in your question, it's not possible to evaluate whether this is likely to help in practice, but it might.