Prove the determinant of the $n \times n$ matrix (whose $(i,j)$-entry is $p_i^{p_j}$) is positive, where $p_1=2,p_2=3,...$ are the smallest $n$ primes.

I know one way to do it is using generalized Vandermonde determinant, but that method has nothing has to do with primes. We don't need the $p_i$'s to be prime, only $\{p_i\}$ is an increasing sequence of natural numbers.

So my question is: is there a way to solve it using properties of primes ?

The hint given is: Induction on $n$, replacing $p_n$ in last row by $x$, and show the determiant function in $x$ has at most $n-1$ roots.


Let $\;a_1,a_2,\dots,a_n\;$ be any increasing sequence of positive real numbers, and let $\;e_1,e_2,\dots,e_n\;$ be any increasing sequence of non-negative integers. The Generalized Vandermonde matrix A := $\;\{a_i^{e_j}\}^n_{i,j=1}\;$ has a determinant which is a polynomial in the $\;a_i.\;$ It has factors $\;\prod_{\le i<j\le n}(a_j - a_i)\;$ and $\;(\prod_{i=1}^n a_i)^{e_1}\;$ and the remaining factor is a polynomial with positive integer coefficients. The second factor is easy because $\;a_i^{e_1}\;$ divides every $\;a_i^{e_j}\;$ since $\;e_1<e_j\;$ by assumption. The first factor is the standard Vandermonde determinant because if $\;a_i=a_j\;$ then the determinant vanishes. There is a proof that the remaining factor has positive coefficients in O. H. Mitchell, Note on determinants of powers, Amer. J. Math. 4(1881), 341–344 but I don't really understand the notation and ideas it is based on. There may be other published proofs.

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    $\begingroup$ Is there any easy proof the remainder has positive integer coefficients? Or does it amount to showing that these coefficients count the # of SSYTs with a given weight? $\endgroup$ – Mike Earnest May 4 '18 at 17:30
  • $\begingroup$ same question as Mike. $\endgroup$ – Kyle YIP May 5 '18 at 11:37
  • $\begingroup$ @MikeEarnest I don't know of any easy proofs. You mention #SSYTs so you probably know a lot more about them than I do. Do you have references for this idea? Perhaps submit your own answer? $\endgroup$ – Somos May 5 '18 at 11:41
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    $\begingroup$ @KyleYIP See Schur polynomials. The ratio between your determinant and the usual Vandermonde determinant is a well studied symmetric function, and their coefficients can be proved to count a discrete object so are positive integers. I think this is hard to prove and not the way you are expected to do this problem. I’ll think more about how to apply the hint. It probably doesn’t have to do with primes at all. $\endgroup$ – Mike Earnest May 5 '18 at 18:00

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