$ y=(x-1)^n \; \; \ldots (i) \\ \; \\ \; \\ Differentiating \; w.r.t. \; x, \\ \; \\ \therefore y_1 \; = \; n(x-1)^{n-1} \; = \; n\dfrac{(x-1)^n}{(x-1)} \; = \; \dfrac{ny}{(x-1)} \; \; \ldots \Big( from \; (i) \Big) \\ \; \\ \; \\ Differentiating \; again \; w.r.t. \; x, \\ \; \\ \therefore y_2 \; = \; n(n-1)(x-1)^{n-2} \; = \; n(n-1)\dfrac{(x-1)^n}{(x-1)^2} \; = \; n(n-1) \dfrac{y}{(x-1)^2} \\ \; \\ \; \\ Differentiating \; again \; w.r.t. \; x, \\ \; \\ \therefore y_3 \; = \; n(n-1)(n-2)(x-1)^{n-3} \; = \; n(n-1)(n-2)\dfrac{(x-1)^n}{(x-1)^3} \; = \; n(n-1)(n-2) \dfrac{y}{(x-1)^3} \\ \; \\ \; \\ \; \\ \therefore y+ \dfrac{y_1}{1!} + \dfrac{y_2}{2!} + \dfrac{y_3}{3!}+\ldots+\dfrac{y_n}{n!} \; = \; \\ y+ \dfrac{ny}{(x-1)} + \dfrac{n(n-1)y}{(x-1)^2} + \dfrac{n(n-1)(n-2)y}{(x-1)^3} + \ldots + \dfrac{n(n-1)(n-2)\ldots 3 \cdot 2 \cdot 1 \; y}{(x-1)^n} \\ \; \\ \; \\ = y \bigg[ 1 + \; ^nC_1 \dfrac{1}{(x-1)} + \; ^nC_2 \dfrac{1}{(x-1)^2} + \; \ldots + \dfrac{1}{(x-1)^n} \bigg] \\ \; \\ \; \\ \; \\ \therefore y+ \dfrac{y_1}{1!} + \dfrac{y_2}{2!} + \dfrac{y_3}{3!}+\ldots+\dfrac{y_n}{n!} \; = \; y \bigg[ 1 + \dfrac{1}{(x-1)} \bigg]^n \\ \; \\ \; \\ = (x-1)^n \Big[ \dfrac{x-1+1}{(x-1)} \Big] \; \; \; \ldots By \; Binomial \; Expansion \\ \; \\ \; \\ \;\\ \therefore y+ \dfrac{y_1}{1!} + \dfrac{y_2}{2!} + \dfrac{y_3}{3!}+\ldots+\dfrac{y_n}{n!} \; = \; (x-1)^n \dfrac{x^n}{(x-1)^n} \; = \; x^n \; \; \; \; Hence \; Proved. $