I would like to prove the following inequality. It arises from my study of random matrices.
I have verified the inequality for $q\in \{0.01,0.02, \ldots, 0.99\}$ and $1\le n\le 100$.


Let $n$ be any positive integer and $0\le q\le 1$. Then the following inequality is true.
$$\sum\_{k=0}^n(-1)^k\binom{n}{k}(q^k-q^n)^n\le (1-q^n)^n.$$


I suspect that the Wilf-Zeilberger method applies here. But I can't get it to work.



Combinatorial inequality involving alternating signs
Mark each box of an $n\times n$ table with probability $q$. By inclusion-exclusion the difference RHS-LHS equals to the probability that there exists a full row (with all boxes marked) but there does not exist a full column: that's because for given $k$ rows the probability that (they are full but no column is full) equals $(q^k-q^n)^n$.


