= Solution
Use positive strikes, the natural real-power domain of this price curve. For any such $K$, the stock-minus-call payoff is
$$
S_1-(S_1-K)^+=\min(S_1,K)>0
$$
almost surely, since $S_1>0$. Therefore the call price must be strictly below the <stock> price $S_0=1$: if $C(K)\geq1$, buying <stock> and selling the call has nonpositive initial cost and strictly positive terminal payoff. Also a negative call price is an immediate <arbitrage> by buying the call.
For $p<0$, $1+K^p>K^p$ and raising to the negative power $1/p$ reverses the inequality, giving $(1+K^p)^{1/p}<K$ and $C(K)<0$. The expression is undefined at $p=0$. For $0<p<1$, strict <concavity> of the power implies $(1+K)^p<1+K^p$, whence
$$
C(K)=(1+K^p)^{1/p}-K>1.
$$
At $p=1$, $C(K)=1$. Every defined case with $p\leq1$ therefore violates the necessary no-arbitrage bounds. Consequently
$$
\boxed{p>1.}
$$
This argument does not require a dense family of strikes; even one positive-strike call gives the contradiction. The strict stock-minus-call payoff explains why the borderline $p=1$ is also excluded.
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