Let $p_{k}(n)$ be a number of partitions of $n$ that into parts not greater than $k$.
$p_{k}(n)=p_{k-1}(n)+p_{k}(n-k)$
i will prove this partition recurrence bu using Generating Functions of Integer partitions. In fact i have an idea but i am not sure whether it is true or not.we can see that;
$$ P(\leq k)=P(=k)+P(\leq k-1) $$ So we have $$ \begin{aligned} &\frac{1}{(1-q)\left(1-q^{2}\right)\dots\left(1-q^{k}\right)}=\frac{q^{k}}{(1-q)\left(1-q^{2}\right)\dots\left(1-q^{k}\right)}\\ &+\frac{1}{(1-q)\left(1-q^{2}\right)\dots\left(1-q^{k-1}\right)} \end{aligned} $$
Since this equality is provided,this proof is completed.i dont know my opinion is right or not?can anyone help please? Have i any mistake?
We can also think any different solutions? Maybe from bijective proof and ferrers diagrams?i am also open and another solutions.thanks.
Your argument seems OK but might be more complicated than necessary. You can instead derive the recurrence from first principles by conditioning on whether a partition contains a part of size $k$ or not. If it does, then removing $k$ yields a partition of $n-k$ into parts $\le k$. If it doesn't, then you have a partition of $n$ into parts $\le k-1$.