Operator precedence is at work:

The '-' operator as a unary operator binds to "2" with the second highest precedence. First comes the ** binary operator, which binds more closely than the unary '-'. Last in your formula comes addition.

It really is spelled out like this:

(-(2**4))+1
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Or in other words: Two to the exponent of four equals sixteen, and then made negative. Negative sixteen plus one equals negative fifteen.

perlop is your friend.

It's the same for common algebraic expressions, hand written or in computer form:

1. Exponent: (**)
2. Sign: (unary -)
3. Division and Multiplication: (/ and *)
4. Subtraction and addition: (- and +)

This is not a complete list of operators, of course. For a better list, again, take a look at perlop.

Edit: fixed parens to clarify.

And, indeed, the Exponentiation section of perlop discusses this very example (well, the -2**4 part anyway).

And how does one know this without "memorizing the precedence table"?

The precedence of simple arithmatic operators was set so that one can write polynomials without using parentheses. So just consider the right polynomial and it is easy to derive what the precedence rules are:

-x3+5x2-6x+2
-(x3)+(5(x2))-(6x)+2

-$x**3+5*$x**2-6*$x+2
-($x**3)+(5*($x**2))-(6*$x)+2
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So, using (( to indicate "binds tighter than" such that the tightest-binding operators are to the left below, the above tells us:

   (( * (( + -
** ((
   (( -u
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(Where -u fits in relative to * nearly doesn't matter, but unary operators tend to bind tighter in general and that holds true here so that the actual precedence is ** (( -u +u (( * / (( + - as can be verified at perlop.)

#!/usr/bin/perl
# http://mathforum.org/library/drmath/view/55721.html

# gives bad output
my $x = ( -1 ** 0 );
print "$x\n";

# correct
my $y = ((-1)**0);
print "$y\n";
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