$$ \text{let number of prodcuts be t}: (x-a)(x-b)(x-c)\dots(x-z)=\sum_{i=0}^{i=t} (-1)^{i+2}x^{t-i}\cdot(\text{lists of length i})\text{ let this summation be called r1} $$ proof by induction: for any number of products, if they can be written in the form of r1 then the result multiplied by the next product also can be written in the form of r1: $$ (x-a)(x-b)\dots(x-y)=\sum_{i=0}^{i=t} (-1)^{i+2}x^{t-i}\cdot(\text{lists of length i}) $$ $$ \text{thus }\sum_{i=0}^{i=t} (-1)^{i+2}x^{t-i}\cdot(\text{lists of length i})\cdot (x-z) $$ $$ =\sum_{i=0}^{i=t}(-1)^{i+2}x^{t-i+1}(\text{lists of length i})+\sum_{i=0}^{i=t}-(-1)^{i+2}x^{t-i}(\text{lists of length i})\cdot z $$ First summation can be written as: $$ =(-1)^2x^{t+1}(\text{lists of length 0})+\sum_{i=1}^{i=t}(-1)^{i+2}x^{t-i+1}(\text{lists of length i}) $$ $$ =(-1)^2x^{t+1}(\text{lists of length 0})+\sum_{i=0}^{i=t-1}-(-1)^{i+2}x^{t-i}(\text{lists of length i+1}) $$
Second summation can be written as: $$ -(-1)^{t+2}x^{0}(\text{lists of length t})\cdot z +\sum_{i=0}^{i=t-1}-(-1)^{i+2}x^{t-i}(\text{lists of length i})\cdot z $$ Adding both summations: $$ =(-1)^2x^{t+1}(\text{lists of length 0})+\sum_{i=0}^{i=t-1}x^{t-i}-(-1)^{i+2}[(\text{lists of length i+1})+(\text{lists of length i})\cdot z ]-(-1)^{t+2}x^{0}(\text{lists of length t})\cdot z $$ $$ =(-1)^2x^{t+1}(\text{lists of length 0}) +\sum_{i=0}^{i=t-1}x^{t-i}-(-1)^{i+2}[(\text{lists of length i+1 including z})+(\text{lists of length i})]-(-1)^{t+2}x^{0}(\text{lists of length t including z thus the whole string}) $$ $$ =\sum_{i=0}^{i=t} (-1)^{i+2}x^{t-i}\cdot(\text{lists of length i}) $$ $$ \text{x-a trivially follow r1 thus the theorem is correct} $$ Is this a valid proof? if it is then how can I make it clearer and improve upon it( I'm going through a discrete math book to learn proofs) and if not how can I correct it?.