Not understanding elliptic curve scalar multiplication to produce Ethereum address

Question

This is the equation

Public key = Private key * G

Here,

 Private key = 0000000000000000000000000000000000000000000000000000000000000001

 G = (0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798,
 0x483ADA7726A3C4655DA4FBFC0E1108A8FD17B448A68554199C47D08FFB10D4B8) 

As Private key =1, there is no iteration on the curve. So value of Public key should be G

But the reality is

Public key = 0x7E5F4552091A69125d5DfCb7b8C2659029395Bdf

NB: Well none Ethereum Private key and Public key

Is there anybody who could make me understand how?

Could you help me showing any math how

Private key = 0000000000000000000000000000000000000000000000000000000000000001 is converted into 0x7E5F4552091A69125d5DfCb7b8C2659029395Bdf as wallet address?

Answer 1

From Yellow Paper of Ethereum

For a given private key, $p_r$ , the Ethereum address $A(p_r )$ (a 160-bit value) to which it corresponds is defined as the right most 160-bits of the Keccak hash of the corresponding ECDSA public key: $$ A(p_r ) = B_{96\ldots255}(\textit{KEC}(\textit{ECDSAPUBKEY}(p_r )$$

• first find the public key in the ECC using the private key $k$

  $$PublicKey = k \cdot G= [k]G $$ set $k=1$ and get $PublicKey = G$

• Now, we need to use Keccak-256 with $x$ and $y$ coordinate is concatanated, we can use an online tool, Never Use for real keys online

• we get

  c0a6c424ac7157ae408398df7e5f4552091a69125d5dfcb7b8c2659029395bdf
  

• Take the rightmost 160-bit from the result or 40-hex from already hex encoded result of hashing.

  0x7E5F4552091A69125D5DFCB7B8C2659029395BDF
  

• Now, we have arrived at your Ethereum address.