Showing that the distribution of record times $(\tau_k)_{k\geq 1}$ doesn't depend on the distribution, $F$, of the records $X_i$

I would show it inductively. For $n=1$ is trivial that $\tau_1$ does not depend on $F$ (simply because $\tau_1=1$.) Next, distribution of $\tau_2$ doesnt depend on $F$:

$$P(\tau_2=k)=P(X_2, \dots, X_{k-1}<X_1<X_k)=\frac{1}{k(k-1)},$$ because $X_i$ are iid and this is simply showing how many ways there is to order variables $X_1, \dots, X_k$ in order such that $X_2, \dots, X_{k-1}<X_1<X_k$ holds. Therefore, distribution of $\tau_2$ is deterministic not depending on $F$.

Now inductively you do the same, $$P(\tau_n=k)=\sum_{i=1}^{k-1} P(\tau_{n-1}=i)P(\tau_n=k\mid \tau_{n-1}=i). $$

Now only thing we need to prove is that $ P(\tau_n=k\mid \tau_{n-1}=i) $ for all $i<k$, doesn't depend on $F$. Then, every element in the sum doestn depend on $F$ which is what we wanted. But

$$ P(\tau_n=k\mid \tau_{n-1}=i)=P[X_k\geq X_i\geq(X_1, \dots, X_{k-1}) \mid X_i\geq(X_1, \dots, X_{i-1})]. $$ This again doesnt depend on $F$, because this is equal to $\frac{1}{k}\cdot (\frac{i-1}{i}\dots\frac{k-2}{k-1})$.

Maybe better explanation: Define $R_i^n$ as the order statistic of $X_i$ in $X_1, \dots, X_n$. I.e. $R_i^n=1$ if $X_i$ is the smallest out of all $X_1, \dots, X_n$.

First observation: $R_i^n$ doesn't depend on $F$. That's because we are looking only for the orders, not on magnitude of each $X_i$.

Second observation: $\tau_k$ can be rewritten using only $R_i^n$. intuitively, thats because we only use $X_i$ is bigger or smaller than $X_j$ in the definition of $\tau$.