2020/015) Find integers n and k such that $n!+8 = 2^k$

Firstly $n\lt 6$ because if $n >= 6$ then is is divisible by $2^4$ ( 2 comes once in 2 twice in 4 and once is $6$ so at least 4 times and $2^4=16$.

we can show it in another way that $6!=720 = 16 * 45$
so $n>6$ $n!$ shall be divisible by 16 so say value is 16m

now $n!+8 = 16m+8 = 8(2m+1)$ and as it is product of 8 and an odd number greater than 1 so it cannot be power or 2 .

Further n cannot be less than 4 because 8 has to  to factor of n

so we need to check for n = 4 and n = 5

no $4!+8 = 24 + 8 = 32$ giving n =  4 k = 5

$5! + 8 = 120 + 8 = 128 = 2^7$ giving n =  5 and k = 7