048.js

// The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.

// Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.

function power(n, power) {
    var digits = [1];
    while (power--) {
        var cr = 0;
        for (var i = 0; i < digits.length; i++) {
            var num = n * digits[i] + cr;
            digits[i] = num % 10;
            cr = Math.floor(num/10);
        }
        while (cr > 0) {
            digits.push(cr % 10);
            cr = Math.floor(cr/10);
        }
    }
    return digits.reverse().join('');
}

function sum(a, b) {
    var lena = a.length;
    var lenb = b.length;
    var sum = '', cr = 0, x, y;
    for (var i = 1; (i <= lena || i <= lenb); i++) {
        x = i > lena ? 0 : parseInt( a.charAt(lena - i), 10 );
        y = i > lenb ? 0 : parseInt( b.charAt(lenb - i), 10 );
        sum = ((x + y + cr)%10) + '' + sum;
        cr = Math.floor((x + y + cr)/10);
    }
    if (cr > 0) {
        sum = cr + '' + sum;
    }
    return sum;
}

function selfPowerSeries(n) {
    var result = '';
    while (n--) {
        result = sum(result, power(n+1, n+1));
    }
    return result;
}

console.log( selfPowerSeries(1000) );