Integers are only 32 bits. This means that its max value is 2^31 -1. As you see, for very small numbers, you quickly have a result which can't be represented by an integer anymore. That's why Math.pow uses double.

If you want arbitrary integer precision, use BigInteger.pow. But it's of course less efficient.

I think there's typically hardware support on most modern processors for doing floating-point powers, but not integers. Because of that, for a general power, it's actually faster to do Math.power with a double and then convert it back to an int.

However, in this case there's a faster way to do it for ints. Since you're doing a power of 2, you can just use the bitwise left-shift operator instead:

int levelMeal = 5*(1<<(level-1));

As Rhymoid pointed out in his comment, that expression can be further simplified to remove the 1:

int levelMeal = 5<<(level-1);

If your inputs are int and the output is always expected to be int, then you're dealing with 32-bit numbers. It's more efficient to write your own function to handle this multiplication rather than using math.Pow. math.Pow, as mentioned in the other answers, expects 64-bit values.

Here's a Benchmark comparison for 15^15 (which approaches the upper limits for 32-bit representation):

// IntPow calculates n to the mth power. Since the result is an int, it is assumed that m is a positive power
func IntPow(n, m int) int {
    if m == 0 {
        return 1
    }

    if m == 1 {
        return n
    }

    result := n
    for i := 2; i <= m; i++ {
        result *= n
    }
    return result
}

// MathPow calculates n to the mth power with the math.Pow() function
func MathPow(n, m int) int {
    return int(math.Pow(float64(n), float64(m)))
}

The result:

go test -cpu=1 -bench=.
goos: darwin
goarch: amd64
pkg: pow
BenchmarkIntPow15   195415786            6.06 ns/op
BenchmarkMathPow15  40776524            27.8 ns/op

I believe the best solution is that you should write your own function similar to IntPow(m, n int) shown above. My benchmarks show that it runs more than 4x faster on a single CPU core compared to using math.Pow.

No, there's no possibility of rounding error caused by the conversion to double. double can exactly represent all integers which fall in the domain of the power function.