For all positive integer $k \ge 2016, n = k!+k$ works.

We use the formula $v_p(n!) = \frac{n - s_p(n)}{p - 1}$ if $s_p(n)$ is the sum of the digits of $n$ in base $p$. Lemma 1. $s_p(x^2) \le s_p(x)^2$. Proof: Let $x = a_1p^{e_1} + a_2p^{e_2} + \ldots + a_kp^{e_k}$, with $\sum a_i = s_p(x)$. Then $x^2 = \sum a_ia_jp^{e_i + e_j}$. We can simplify each $a_ia_j$ into a base $p$ expansion which reduces its sum of digits. And then this is just equivalent to adding a bunch of numbers in base $p$, which may or may not have a bunch of carries which only reduce the digit sum further. And the initial "digit sum" right after squaring was $s_p(x)^2$. In fact, $s_p(xy) \le s_p(x)s_p(y)$. Lemma 2. Given any positive integer $k$, there are infinitely many integers $x$ with $s_p(x) \ge k$ for all primes $p$. We delegate the proof of this to the bottom. Then take one of the integers as in Lemma 2, for $k = 2015$. For all primes $p$ we have $v_p\left(n!^{n + 2015}\right) = (n + 2015)\left(\frac{n - s_p(n)}{p - 1}\right) \le (n + s_p(n))\left(\frac{n - s_p(n)}{p - 1}\right) = \frac{n^2 - s_p(n)^2}{p - 1} \le \frac{n^2 - s_p(n^2)}{p - 1} = v_p((n^2)!)$, and thus this $n$ works. Proof of Lemma 2: We use very bad asymptotics. Essentially look at how many integers at most $n$ fail for each prime $p$ (we only need to consider $p \le n$, since if $p > n$ then the number $1 \le x \le n$ is just $x_p$, which for $x \ge k$ wins, and the $k - 1$ which fail are just an extra constant wrt $n$). So the guys who fail are at most $n \le p^y$ for $\log_p n \le y < \log_p n + 1$ an integer. And if we use any of many combinatorial techniques, the amount is equivalent to solving $a_1 + a_2 + \ldots + a_y \le k$ for nonnegative integers $a_1, \ldots, a_y$, and counting the ways (which is clearly an upper bound), which turns out to be (by aforementioned ways) at most $c_ky^k$, for some constant $c_k$ independent of $n$ and $p$. And $y \le 2\log_p n$, so there are at most $d_k(\log n)^k\frac{1}{(\log p)^k}$ ways for a constant $d_k$ independent of $n, p$. Now sum over all primes $p \le n$. There is a constant $c$ with $\pi(x) \le c\frac{x}{\log x}$ for all $x \ge 2$, and since $f(x) = \frac{1}{(\log x)^k}$ is decreasing the sum is clearly bounded from above by $d_k(\log n)^k\left(\sum_{x = 2}^{c\frac{n}{\log n}} \frac{1}{(\log x)^k}\right)$. And now $\sum_{x = 2}^a \frac{1}{(\log x)^k}$ is less than $e_k\frac{a}{(\log a)^k}$ for some constant $e_k$ independent of $x, a$ (this is easy to see in many ways). So our total count is bounded above by $d_k(\log n)^k\left(e_k\frac{c\frac{n}{\log n}}{(\log n - \log \log n)^k}\right) << \frac{n}{2}$ for large enough $n$. Thus we win; as $n \rightarrow \infty$ we get lots of guys who work; at least half of the integers work after a certain point, in fact. The same is true for any fraction. The last statement is clear if we note that $(\log n)^{\frac{k - 1}{k}} + \log \log n << \log n$ asymptotically.