Suppose you need 2^0 mod 4. The answer is 1 and you'll find 2^phi(4)=0

The function for computing what you want is the following (try to understand it as an exercise)

int pow_mod (int a, int n, int mod)
{
    int ans = 1;

    for (; n; n >>= 1)
    {
        if (n & 1)
            ans = (ans * 1ll * a) % mod;

        a = (a * 1ll * a) % mod;
    }

    return ans;
}

In the case where n is a big integer, you only need to convert it to binary and the apply the function above. I hope this can help you.

NOTE: In the case that you don't understand the above function, just ask me