Statement $P(n) : a^n +b^n = c^n +d^n $
here we have to prove $P(n)$ is true for all $n \in N$
So, $P(1) : a+b = c+d$
hence $P(1)$ is true (given in the question)
Now, assume $P(m) : a^m +b^m = c^m + d^m $ is true for all $m \leq k$ for some $m,n \in N$.................. (1)
Now, with the help of the assumption (1) we have to prove $a^{k+1} + b^{k+1}= c^{k+1} + d^{k+1}$
Now,
$(a^k + b^k) (a+b) = (c^k +d^k) \ (c+d)$ ( since we have assumed in $1$ that $a^m +b^m = c^m+d^m$ for all $m \leq k$ )
$\Rightarrow$ $a^{k+1} + ab^k + a^k b + b^{k+1} = c^{k+1} + cd^k + c^k d + d^{k+1}$
$\Rightarrow$ $a^{k+1}+ b^{k+1} +ab( b^{k-1} + a^{k-1}) = c^{k+1} + d^{k+1} + cd(d^{k-1} + c^{k-1})$
$\Rightarrow$ $a^{k+1}+ b^{k+1} = c^{k+1} +d^{k+1}$ ( Here $ab = cd$ (given) and $(a^{k-1} + b^{k-1}) = (c^{k-1} +d^{k-1})$ (from assumption (1))
So, $P(n)$ is true for all $n \in N$