The solution is
#y=c_0sum_{n=0}^infty{x^n}/{n!}=c_0e^x#,
where #c_0# is any constant.
Let us look at some details.
Let
#y=sum_{n=0}^infty c_n x^n#
#y'=sum_{n=1}^infty nc_n x^{n-1}=sum_{n=0}^infty(n+1)c_{n+1}x^n#
So, we can rewrite #y'-y=0# as
#sum_{n=0}^infty (n+1)c_{n+1} x^n-sum_{n=0}^infty c_n x^n=0#
by combining the summations,
#Rightarrow sum_{n=0}^infty[(n+1)c_{n+1}-c_n]x^n=0#
so, we have
#(n+1)c_{n+1}-c_n=0 Rightarrow c_{n+1}=1/{n+1}c_n#
Let us observe the first few terms.
#c_1=1/1c_0=1/{1!}c_0#
#c_2=1/2c_1=1/{2}cdot1/{1!}c_0=1/{2!}c_0#
#c_3=1/3c_2=1/3cdot1/{2!}c_0=1/{3!}c_0#
.
.
.
#c_n=1/{n!}c_0#
Hence, the solution is
#y=sum_{n=0}^infty1/{n!}c_0x^n=c_0sum_{n=0}^infty{x^n}/{n!}=c_0e^x#,
where #c_0# is any constant.