= Solution
Put $h=H^2(P_0,P_1)$ with the paper's normalized <Hellinger distance>. The <Hellinger affinity> is
$$
A(P_0,P_1)=\int\sqrt{p_0p_1}\,d\mu=1-h,\qquad0\leq A(P_0,P_1)\leq1.
$$
The last bound follows from the <Cauchy-Schwarz inequality> and the unit integrals of the two <probability density functions>. Using a common dominating <probability measure>, for example $(P_0+P_1)/2$, makes the product calculation immediate. The <probability density functions> of the <product measures> factor, and <Tonelli's theorem> gives
$$
A(P_0^n,P_1^n)
=\int\prod_{i=1}^n\sqrt{p_0(x_i)p_1(x_i)}\,d\mu^{\otimes n}(\mathbf x)
=A(P_0,P_1)^n.
$$
The <Squared Hellinger distance> is independent of the choice of common dominating measure, so this calculation applies to the original distributions. We obtain the exact identity and then apply <Bernoulli's inequality>:
$$
\boxed{H^2(P_0^n,P_1^n)=1-(1-h)^n\leq nh=nH^2(P_0,P_1).}
$$
For completeness, the scalar bound follows by induction: $(1-h)^1=1-h$, and multiplying $(1-h)^n\geq1-nh$ by the nonnegative $1-h$ gives $(1-h)^{n+1}\geq1-(n+1)h+nh^2\geq1-(n+1)h$.
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