$T:P_nlongrightarrow P_n : Tp(t)=frac{d}{dt}p(t)text{ Find the norm of the operator}$

Let $$P_n:=bigg{p:[0,1]rightarrowmathbb R :deg(p)le nbigg}$$
And define the norm $$lVert p(t)rVert=max_{0le kle n}lvert a_krvert text{ where $p(t)=a_nt^n+…+a_1t+a_0$}$$

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We define a linear operator: $$T:P_nlongrightarrow P_n :$$
$$Tp(t)=frac{d}{dt}p(t)$$
Find the ||T|| norm.

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Ok so my thoughts so far are:
$$p'(t)=color{black}{underbrace{na_n}_{b_n}}t^{n-1}color{black}{underbrace{(n-1)a_{n-1}}_{b_{n-1}}}t^{n-2}+…+color{black}{underbrace{1cdot a_1}_{b_1}}+color{black}{underbrace{0cdot a_0}_{b_0}}$$
$$text{Hence}:b_k=kcdot a_k,k=0,1,2,…,n$$
$$text{So: } lVert Tp(t)rVert=lVert p'(t)rVert=max_{0le kle n}lvert b_krvert=max_{0le kle n}lvert ka_krvertle nmax_{0le kle n}lvert a_krvert$$
$$=nlVert prVert $$
$$text{Thus, }quad bbox[3px,border:2px solid red] {lVert TprVert le nlVert prVert } qquad (1)$$
$$text{Let }p_0(t)=1cdot t^nRightarrowlVert p_0rVert =1$$
$$text{By the definition of the norm we get :}$$
$$lVert TrVert =sup_{lVert prVert=1}lVert TprVertge lVert Tp_0rVert=lVert p_0’rVert=lVert ncdot t^{n-1}rVert=nquad (2)$$
$$text{Hence : by (1),(2) it implies that : }lVert TrVert=n. $$

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Thoughts on that?
Thank you.

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