The general technique for this involves storing a large number as a sequence of, say, nine digit numbers. Multiplication is performed just the same as manual multiplication is, except that instead of one digit at a time, it's nine digits at a time. The digits that won't fit are carried over to the next word.
A cleverer technique for calculating large powers involves a full bigint x bigint routine and working how to multiply partial products to reduce the number of multiplications (e.g., 2^64 = (2^32)^2 = ((2^16)^2)^2 = ((((2^8)^2)^2)^2 = ((((((2^4)^2)^2)^2)^2 = (((((((2^2)^2)^2)^2)^2)^2 -- reducing the number of multiplications from 64 to 6 -- a 10x reduction).
The only real trick is printing leading zeroes on every number except the first one. My easy solution is below.
#define MOD 100000000 /* 10^9 might have been a better choice */
int ans[15000];
char out[1000020];
int maxans = 0;
int main() {
FILE *fin = fopen ("cruel1.in", "r");
FILE *fout = fopen ("cruel1.out", "w");
int n, p, i, j, k, carry;
char *outp;
long long partial;
fscanf (fin, "%d %d", &n, &p);
ans[0] = 1;
for (k = 0; k < p; k++) {
/* multiply 'ans' by n */
carry = 0;
for (i = 0; i <= maxans || carry; i++) {
partial = n * (long long) ans[i] + carry;
if (partial >= MOD) {
ans[i] = partial % MOD;
carry = partial / MOD;
} else {
ans[i] = partial;
carry = 0;
}
if (ans[i] && i > maxans) maxans = i;
}
}
/* print answer */
outp = out;
for (i = maxans; i>=0; i--) {
if (i == maxans) sprintf (outp, "%d", ans[i]);
else sprintf (outp, "%08.8d", ans[i]);
while (*outp) outp++;
}
for (i = 0; out[i]; i+= 70)
fprintf(fout, "%.70s\n", &out[i]);
exit (0);
}