Let $H_n$ be the nth harmonic number, then $$H_n=\sum_{i=1}^n\frac{1}{i}=\frac{1}{n!}\sum_{i=1}^n\frac{n!}{i}$$ Let $v_2(n!)=a$. Note that if we can show $$v_2\left(\sum_{i=1}^n\frac{n!}{i}\right)<v_2(n!)$$ we're done.
Since $$v_2\left(\sum_{i=1}^n\frac{n!}{i}\right)=v_2\left(\frac{n!}{2^a}\right)=v_2(n!)-a<v_2(n!)$$ That completes the proof. The thing that I'm not sure of is the generalization of this lemma that I've used $$v_p(a+b)=v_p(b)$$ whenever $v_p(b)<v_p(a)$. Its generalization is as follows,
Let $(a_n)$ be a sequence of positive integers s.t. $v_p(a_n)>...>v_p(a_1)$ then $$v_p\left(\sum_{i=1}^na_i\right)=v_p(a_1)$$ But again I'm not sure whether it's true or not.