Inspired by a task for Programming 101, here's a challenge that hopefully isn't too easy.
n >= 1.n lines of asterisks, where every new line has one asterisk more than the line before, and starting with one asterisk in the first line.* ** *** **** ***** This is code-golf, so shortest answer in bytes wins.
s(0)->"";s(N)->s(N-1)++string:copies("*",N)++"~n". s(0)->""; % If the current item is 0, return the empty string. s(N)-> % Otherwise: s(N-1) % Return the N-1 string. ++string:copies("*",N). % Append "*" N times. ++"~n" % Append a newline. 1.."$args"|%{$a+="*";$a} 
#include<ios> int f(int n){for(n&&f(n-1);~n;putchar(n--?42:10));} Including the #include in the byte count.
f(n){; int is implied as both parameter and return type. \$\endgroup\$ L,R"*"€*BFn A function that returns the string given an argument \$n\$.
A full program is 17 bytes:
y:? Ey,x:"*",*i,O L, ; Define "anonymous" function ; It takes one argument, e.g 5 R ; Range STACK = [[1 2 3 4 5]] "*" ; Push "*" STACK = [[1 2 3 4 5] "*"] € ; For each i: * ; Repeat "*" i times STACK = [["*" "**" "***" "****" "*****"]] BF ; Flatten STACK = ["*" "**" "***" "****" "*****"] n ; Join by newlines STACK = ['*\n**\n***\n****\n*****'] ; Return the top value y:? ; y = input() Ey, ; for i in range(y): x:"*", ; x = "*" *i, ; x *= i O ; print(x) 
d, \$9\log_{256}(96)\approx\$ 7.41 bytesR{1+'**ZK Attempt This Online! The slight misallignment on the third line of output is just ATO's formatting.
Explanation:
R{1+'**ZK # Implicit input R{ # Loop through range(0, input) 1+ # Add one * # Multiply this '* # With the string "*" ZK # Print # The d flag stops the implicit print at the end 
time {puts [append s *]} $n time {puts [string repe * [incr i]]} $n while {[incr i]<=$n} {puts [string repe * $i]} 
let s={(1...$0).map{print({String.init}()("*",$0))}} 