Is this a bug

[Starpower_Power]

a=1
for i in range(1,5)
a=i*a^i
end

now result is zero
is this kind of bug

[Oscar_Smith]

Discussion about integer overflow

Yes, it is a limitation but I would not call it big since we could use BigInt if Int64 is not enough and an overflow is something we can catch. I find more worrying this: > julia> 1 / 10^20000 > Inf I’d rather have an overflow here than Inf, or at least an Infinitesimal value… Maybe something we could ask for future versions of Julia.