Proof of $ r(I)^e\subset r(I^e)$ in ring theory with extension and radical 
Let $f:A\to B$ be a ring homomorphism and $I$ be an ideal of $A$. Then prove that $\displaystyle r(I)^e\subset r(I^e)$, where $I^e$ denotes the extension of the ideal $I$ and $r(I)$ denotes the radical of the ideal $I$.

Let $x\in r(I)^e=Bf(r(I))$. Then $x=yz$ for some $y\in B$ and $z\in f(r(I))$. So $z=f(z_1)$ for some $z_1\in r(I)\implies z_1^n\in I$ for some $n>0$. Then $z^n=f(z)^n=f(z^n)\in f(I).$
Now $x^n=y^nz^n$ where $y^n\in B$ and $z^n\in f(I)$. So $x^n\in I^e\implies x\in r(I^e)$.
Is the proof correct ? Please check and if there are some mistake then give hint how to proceed.

As user26857 pointed out, $x=yz$ is not right. Remember that $I^e=\langle f(I)\rangle$, so it should be instead $x=\sum_{i=0}^n b_if(u_i)$ for some $n\in \Bbb Z^+$, where $b_i\in B$ and $u_i\in \sqrt{I}$. 
Now, as $u_i\in \sqrt{I}$, then there is $k_i\in \Bbb Z^+$ such that $u_i^{k_i}\in I$. Therefore, $$\bigl(b_if(u_i)\bigr)^{k_i}=b_i^{k_i}f(u_i^{k_i})\in I^e$$ $$\implies b_if(u_i)\in \sqrt{I^e}.$$
Hence, as $\sqrt{I^e}$ is an ideal of $B$, it follows that $x\in \sqrt{I^e}$.

Alternatively, you can use the multinomial theorem to $\Bigl(\sum_{i=0}^n b_if(u_i)\Bigr)^k$, where $k=k_1+\cdots +k_n$.
