Let $a = \sqrt[2015]{2015}$. Note that $a > 1$ since $2015 > 1$.

We have $a_1 = a$ and $a_{n+1} = a^{a_n}$ for $n \ge 1$.

Let us compute the first few terms:

$a_1 = a$

$a_2 = a^{a_1} = a^a$

$a_3 = a^{a_2} = a^{a^a}$

and so on.

Let us try to estimate $a_n$ and see if it can ever reach $2015$.

First, note that $a = 2015^{1/2015}$.

Let us compute $a$ numerically:

$\ln a = \frac{1}{2015} \ln 2015$

$\ln 2015 \approx 7.616$ (since $\ln 2000 \approx 7.601$)

So $\ln a \approx \frac{7.616}{2015} \approx 0.00378$

Thus, $a \approx e^{0.00378} \approx 1.00379$

So $a$ is just slightly greater than $1$.

Now, $a_2 = a^a \approx (1.00379)^{1.00379}$

Let us compute $\ln a_2 = a \ln a \approx 1.00379 \times 0.00378 \approx 0.00379$

So $a_2 \approx e^{0.00379} \approx 1.00380$

Similarly, $a_3 = a^{a_2} \approx (1.00379)^{1.00380}$

$\ln a_3 = a_2 \ln a \approx 1.00380 \times 0.00378 \approx 0.00380$

So $a_3 \approx e^{0.00380} \approx 1.00381$

We see that $a_n$ increases extremely slowly.

Let us try to estimate how large $a_n$ can get.

Suppose $a_n \approx 1 + c_n$, where $c_n$ is very small.

We see that $a_{n+1} = a^{a_n} = e^{a_n \ln a} \approx e^{(1 + c_n) \cdot c_1}$, where $c_1 = \ln a \approx 0.00378$.

So $a_{n+1} \approx e^{c_1 + c_n c_1} \approx 1 + c_1 + c_n c_1$ (using $e^x \approx 1 + x$ for small $x$).

Thus, $c_{n+1} \approx c_1 + c_n c_1$

Let us try to see how $c_n$ grows:

Let $c_1 = \ln a \approx 0.00378$

$c_2 \approx c_1 + c_1^2 \approx 0.00378 + (0.00378)^2 \approx 0.00378 + 0.0000143 \approx 0.003794$

$c_3 \approx c_1 + c_2 c_1 \approx 0.00378 + 0.003794 \times 0.00378 \approx 0.00378 + 0.00001435 \approx 0.0037944$

So $c_n$ increases by about $0.000014$ each time, which is extremely slow.

To reach $a_n \ge 2015$, we need $a_n \ge 2015$, i.e., $c_n \ge 2014$.

But $c_n$ increases by about $0.000014$ per step, starting from $0.00378$.

So the number of steps required is roughly $\frac{2014}{0.000014} \approx 144,571,429$ steps.

But actually, the increment per step increases slightly, but still, the growth is extremely slow.

Therefore, for all practical purposes, $a_n$ will never reach $2015$ for any reasonable $n$.

Thus, the answer is:

No, there does not exist a positive integer $n$ such that $a_n \ge 2015$.