Is every Wolstenholme number greater than or equal to its index?

Background

This question was inspired by this code golf post, and I've taken some of this background explanation from there.

Consider the generalised harmonic numbers of order 2:

$$H_{n,2} = \sum^n_{k=1} \frac 1 {k^2}$$

This sequence begins:

$$1, \frac 5 4, \frac {49} {36}, \frac {205} {144}, \dots\ \text{and converges to } \frac {\pi^2} 6\ \text{as } n \to \infty$$

However, the numerators of this sequence form another sequence known as the Wolstenholme numbers (A007406):

$$1, 5, 49, 205, 5269, 5369, 266681, 1077749, 9778141, ...$$

Question

Let $i$ be the index and $w(i)$ the Wolstenholme number at index $i$.

So $w(2) = 5$, $w(3) = 49$, and so on.

Is it the case that $w(i)$ is always greater than or equal to $i$?

The initial terms suggest the answer is yes, but it might be possible that, when reduced to lowest terms, the fraction producing $w(i)$ collapses down so much that its numerator is less than its index $i$.

Is there an argument to demonstrate that this cannot happen?

One can adapt the proof that the usual harmonic numbers are never integers to show this. Suppose that $k$ is chosen so that $2^k \le n < 2^{k+1}$. Then $2^{2k}$ divides exactly one denominator in the definition of $H_{n,2}$, which implies that $2^{2k}$ divides the denominator of $H_{n,2}$ itself even when reduced to lowest terms (this implication holds for any prime power). Therefore the numerator of $H_{n,2}$ is also at least $2^{2k} > 2^{k-1}n \ge n$ when $k\ge2$, that is, when $n\ge 2^2 = 4$; and the claim can be checked by hand for $1\le n\le 3$.