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There is an exercise in section 4.1 of Hatcher:

Show that an $n$-connected, $n$-dimensional CW-complex is contractible.

I know this can be proved by using the Hurewich theorem from section 4.2 easily (let me know if my argument is wrong):

If $X$ is $n$-connected then $\pi_k(X)~\text{for}~ k=0,\cdots, n$. Since $X$ is an $n$-dimensional CW complex, $H_k(X)=0$ when $k\ge n+1$. By using Hurewich theorem (Theorem 4.32) inductively, we have $\pi_{k}(X)=H_{k}(X)=0$ for $k>n$. Then it follows from the Whitehead theorem that $X$ is contractible.

My question is: since this problem is given before the section for Hurwich theorem, is there an approach by using only the knowledge in section 4.1 of Hatcher?

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    What version of Whitehead are you using? It sounds like you are using the usually corollary (which comes from Hurewicz) of the standard Whitehead theorem. – 2017-02-14
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    @PVAL-inactive The version I am using is theorem 4.5(page 346) of his book, which is proved before the Hurwich theorem(4.32) – 2017-02-14
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    Alright so then what do you need Hurewicz for? You do understand that $\pi_k(X)$ trivial IS the definition of n-connected? – 2017-02-14
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    @PVAL-inactive I was trying to show that $\pi_{k}(X)=H_{k}(X)=0$ for $k>n$. Yes, the condition says $\pi_{k}(X)=0$ for $k\le n$, after showing that all homotopy groups are trivial, I can use the whitehead theorem to prove that $X$ is contractible – 2017-02-14
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    You build the contraction by hand, as in the proof of Whitehead. – 2017-02-15

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Using Contraction Lemma, set $Y=X$, $B=x_0$. By the lemma, any map (including identity of X) should homotopic to the constant map. Thus it is contractible.