Show that if a|c, b|c and gcd(a,b)=1, then ab|c Let $a,b,c \in \mathbb{Z}$ such that $a|c$, $b|c$, gcd$(a,b)=1$. Prove that $ab|c$.
My thoughts so far: By the Unique Factorization Theorem, we can rewrite $a$ as $p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$ where $p_i$'s are primes that make up $a$. Similarly, we can write $b=q_1^{\beta_1}q_2^{\beta_2}...q_k^{\beta_k}$, but with the restriction that $p_i\neq q_j$, $\forall i,j$. Then, by definition of "$|$," we can say that $a|c$ implies $\exists k \in \mathbb{Z}$ such that $k(p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k})=c$. (Similarly for $b$).
From here I am unsure where to go, but it feels like I am only a lemma away from my conclusion.

Any time you see $gcd(a,b)=1$, it can often be helpful to think of Bezout's identity. 
$$
gcd(a,b)=1\Rightarrow \exists \; x,y\in\mathbb{Z}\;s.t\; 1=ax+by
$$
Then we can scale to get
$$
c=cax+cby
$$
And now using that $a\vert c,\;b\vert c \Rightarrow c=ak_1=bk_2$ our identity becomes
$$
c=cax+cby=axbk_2+aybk_1\Rightarrow c=ab(k_2x+k_1y)
$$
