Let $p$ be any prime number and $k$ be any natural number.
$3$$^($$^p$$^-$$^1$$^)$$^($$^p$$^k$$^-$$^1$$^)$$=$$(2+1)$$^($$^p$$^-$$^1$$^)$$^($$^p$$^k$$^-$$^1$$^)$ $\equiv {2^{(p-1)(pk-1)}}+1\pmod {(p-1)}$
$2^{(p-1)(pk-1)}$$+1=$${(1+1)}$$^{(p-1)(pk-1)}$${+1}$$\equiv 2+1=3\pmod {(p-1)}$
That's what I calculated but when I try it on ${k=1}$ ${ p=5}$
$2^{(5-1)(5-1)}$$+1\equiv 1\pmod {(p-1)}$ not $3$.
Where did I make mistake ?
If $w$ denotes the power, your congruences assume that the binomial coefficients $\binom{w}{k}$ are all zero mod $w$ for $1 \le k \le w-1.$ This is true when the exponent $w$ is prime, but your eponent is $(p-1)(pk-1)$ for $p$ a prime. That exponent is not prime except when $p=2,k=2.$ In that one case the exponent is $3.$ Note in that case your congruence says $27 \equiv 3 $ mod $2$ which is right.