Euclid's Lemma and $\gcd(a,b)=1\Rightarrow \gcd(a^n,b^k)= 1$

Your answer to the problem is correct. There is another, very elegant approach using the universal property of $\mathrm{gcd}$, which I learned from this answer. The universal property of $\mathrm{gcd}$ is that $$ x|(a,b)\qquad\Leftrightarrow \qquad x|a\quad\text{and}\quad x|b. $$ Now the proof is as follows. Since $c|a$ we have $(c,d)|a$. Since $d|b$ we have $(c,d)|b$. Hence the universal property of $\mathrm{gcd}$ implies that $(c,d)|(a,b)$. Since $(a,b)=1$ it follows that $(c,d)=1$.

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This approach can also be used to prove the second statement. Note that $$(a,bc)|(ac,bc)$$ because $(a,bc)$ divides $a$ (and hence $ac$) and $bc$. Also, we have that $(ac,bc)/c\ |\ (a,b)$ because $$\frac{(ac,bc)}{c}\left|\frac{ac}{c}\right.\qquad\text{and}\qquad\frac{(ac,bc)}{c}\left|\frac{bc}{c}\right.$$ It follows that $(a,bc)|(a,b)c=c$. Since we also have that $(a,bc)|a$, the universal property we get that $(a,bc)|(a,c)$ and therefore $(a,bc)=1$. The third statement follows from the second with induction on $n$ and $k$.

Hint $\ \ \begin{eqnarray}\rm (c,d)&\mid c\mid a\\ \rm(c,d)&\mid d\mid b\end{eqnarray}\ \Rightarrow\rm\ (c,d)\mid a,b\:\Rightarrow\: (c,d)\mid (a,b)\ \ \ $ QED

EL = Euclid's Lemma $\rm\:(a,b)=1=(a,c)\:\Rightarrow\:(a,bc) = 1,\:$ is a special case of

$$\rm\ (a,b,c) = 1\ \Rightarrow\ (a,b)(a,c) = (a(a,b,c),bc) = (a,bc) $$

Thus iterating EL: $\rm\:\ (a_i,b_j) = 1\:\Rightarrow\:(a_i,\prod b_j)=1\:\Rightarrow\:(\prod a_i,\prod b_j) = 1.$

Specializing $\rm\:a_i = a,\ b_j = b\:$ we deduce $\rm\:(a^n,b^k) = 1.\:$

It looks fine to me. But really, $a = ck$ and $b = dj$, then if $c$ and $d$ had a common factor, $a$ and $b$ would clearly have as well, right?