= Solution
For <number fields>, a <prime ideal> divides the <different ideal> exactly when it is ramified. More precisely, the local <different exponent and tame ramification> relation is $d_{\mathfrak P}\ge e_{\mathfrak P}-1$, with equality for tame ramification; if $e=1$, the finite residue extension is separable and the exponent is zero. One can see the ramification criterion directly from the <trace pairing> modulo the base prime: when $e>1$ its quotient ring has a nonzero nilpotent ideal, whose elements pair to zero since multiplication by a nilpotent has trace zero. When $e=1$ the quotient is a product of finite separable fields and its trace pairing is nondegenerate. Its determinant is a unit exactly in the second case.
Let $M=K_1\cap K_2$. Any rational prime ramifying in $M$ must ramify in each $K_i$, because <ramification indices> multiply in a tower. It would then divide both <field discriminants>, contrary to their coprimality. Thus $M$ is unramified at every finite prime, so $|d_M|=1$. The <Minkowski lower bound for a number-field discriminant> excludes this for a field of degree $n>1$:
$$
\sqrt{|d_M|}\ge\left(\frac\pi4\right)^{r_2}\frac{n^n}{n!}>1.
$$
The last inequality follows already for the weakest signature bound $r_2\le n/2$: the expression starts above one at $n=2$ and increases with $n$. Consequently $M=\mathbb Q$.
For completeness, normality converts this intersection statement into the desired degree statement. Put $E=K_1K_2$. Restriction injects $\operatorname{Gal}(E/K_2)$ into $\operatorname{Gal}(K_1/\mathbb Q)$. The fixed field of its image in $K_1$ is exactly $K_1\cap K_2$, because $E/K_2$ is <Galois>. The <Galois correspondence> therefore gives $[E:K_2]=[K_1:M]=[K_1:\mathbb Q]$. This proves <coprime discriminants imply linear disjointness>:
$$
\boxed{[K_1K_2:\mathbb Q]=[K_1:\mathbb Q][K_2:\mathbb Q].}
$$
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