Proof that the diff-in-diff (wrt sample size) of the expectation of a first-order statistic is positive (Stigler 1961)
Well, you cannot take the derivative of $E(n)$ with respect to $n$, because $n$ is an integer variable.  

More generally, you want to prove a property with respect to $n$. The problem you have is that the corresponding domain is not a convex set: say, for $n$ and $n+1$, $0&lt;\lambda &lt;1$, the value $\lambda n + (1-\lambda) (n+1) = n+1-\lambda$ does not belong to the domain since it is not an integer. But if the domain of the function, with respect to the variable you are interested in, is not a convex set, the concept of convexity/concavity is not even defined (perhaps in higher mathematics they have some more nuanced concept to accommodate that).

To determine whether the inequality you want holds, a brute-force algebraic approach is to examine its sign directly: we have

$$[E(n+2) - E(n+1)] - [E(n+1) - E(n)] =\\= (n+2) \int_0^\infty p (1-F)^{n+2-1} F' \mathrm d p \\- 2 (n+1) \int_0^\infty p (1-F)^{n+1-1} F' \mathrm d p\\ +n \int_0^\infty p (1-F)^{n-1} F' \mathrm d p$$

$$= \int_0^\infty p (1-F)^{n-1} F'\cdot\Big[(n+2)(1-F)^2 - 2(n+1)(1-F)+n\Big] \mathrm d p $$

Doing the algebra

$$(n+2)(1-F)^2 - 2(n+1)(1-F)+n  \\= n(1-F)^2 + 2(1-F)^2 - 2n(1-F) - 2(1-F) +n \\
= n\Big[(1-F)^2 - 2(1-F) +1\Big]  - 2(1-F)F \\
= n [(1-F) -1]^2 - 2(1-F)F = nF^2 -  2(1-F)F = F\Big[nF-2+2F\Big]\\
=F\cdot \Big[(n+2)F-2\big]$$

So we have arrived at

$$[E(n+2) - E(n+1)] - [E(n+1) - E(n)] = \int_0^\infty p (1-F)^n F'\cdot\Big[(n+2)F-2\Big] \mathrm d p$$

Breaking up and manipulating,

$$[E(n+2) - E(n+1)] - [E(n+1) - E(n)] = \\=(n+2)\int_0^\infty p (1-F)^{n+1} F' \mathrm d p \\- \left(\frac{2}{n+1}\right)(n+1)\int_0^\infty p (1-F)^n F'\mathrm d p$$

$$\implies [E(n+2) - E(n+1)] - [E(n+1) - E(n)] = E(n+2)-\frac{2}{n+1}E(n+1)$$

Simplifying $E(n+2)$ from both sides we get

$$-2E(n+1) + E(n) = -\frac{2}{n+1}E(n+1)$$

$$\implies E(n) = \frac {2n}{n+1}E(n+1)$$

and by induction

$$\implies E(n+1) = \frac {2(n+1)}{n+2}E(n+2) \implies E(n+2) = \frac {n+2}{2(n+1)}E(n+1)$$

Therefore

$$[E(n+2) - E(n+1)] - [E(n+1) - E(n)] = $$
$$= \left(\frac {n+2}{2(n+1)} -1\right)E(n+1) - \left( 1- \frac {2n}{n+1}\right)E(n+1) $$
$$= \left(\frac {n+2}{2(n+1)} -1 -  1+ \frac {2n}{n+1}\right)E(n+1)$$

$$=\frac {n+2 - 4(n+1) + 4n}{2(n+1)}E(n+1) = \frac {n-2}{2(n+1)}E(n+1)$$

So $n=1$ is not a search, for $n=2$ we have equality, and for $n\geq 3$ the inequality holds.