How do we prove the continuity of the exponential function restricted to $\mathbb{Q}$? 
Let $M$ be a natural number and, for $p/q\in \mathbb{Q}$, define $M^{p/q}$ as $\sqrt[q]{M^p}$ (forget about $a^x$ when $x$ is not rational). Prove that $f:\mathbb{Q}\to \mathbb{R}$ is continuous.

If we define the exponential function as a series, then, by uniform convergence theory the continuity follows easily. Now, if we don't use that definition and we stay in $\mathbb{Q}$ where the definition is done without using difficult concepts, how can we prove continuity over $\mathbb{Q}$? More precisely, i want to prove that if $\{a_n\}_n\subseteq\mathbb{Q}$ converges to $a$, then $\{M^{a_n}\}_n$ converges to $M^a$
I've tried it and it seems it's not so easy (for me). I hope some of you can give me  a hand, thanks.
 A: For $M\ge 2$ and fixed rational $s$, suppose $r_n \to s^+.$ Write (for later) $M=1+a$ so that $a=M-1\ge 1.$ Then $M^{r_n}-M^s=M^s(M^{r_n-s}-1).$ So we want to show that $M^{r_n-s}-1 \to 0$ as $n \to \infty.$ Since $r_n-s \to 0$ from above, for any $k$ it is eventually less than $1/k.$ So if we show $M^{1/k}-1 \to 0$ we are finished (for the $r_n \to s$ from above case), because we have shown $M^{r_n}-M^s \to 0.$
Now note the inequality
$$M^{1/k}=(1+a)^{1/k}<1+\frac ak, \tag{1}$$
which follows from the power map $f(u)=u^k$ being increasing and the fact that $1+a <(1+\frac ak)^k$ since the first few terms of the latter are $1+a+(k\cdot(k-1)/2)(a/k)^2,$ and remaining terms are positive. Thus since the upper bound $1+\frac a k \to 1,$ and since $g(x)=M^x$ is increasing, we arrive at $M^{1/k}\to 1,$ that is, $M^{1/k}-1 \to 0.$
If $r_n \to s$ from below a similar argument works, applying the reciprocal inequality of $(1),$ namely 
$$(1+a)^{-1/k} >(1+a/k)^{-1}.$$
Note: that $M^x$ is increasing is easily shown; details if anyone wants them can be included.
